This collection offers 20 practice problems on monomials and polynomials, arranged so as to follow the theory step by step. We begin by recognising monomials and reducing them to standard form; we then examine the coefficient, the literal part, the degree and the basic operations. Finally, we turn to the structure of polynomials, to their classification, to the operations on them and to the evaluation of a polynomial at given values.
Each problem comes with a complete solution, in which the rules employed are justified and the conditions required for their application are made explicit. The aim is not merely to arrive at the correct answer, but to understand the structure of the expressions involved and the significance of every step.
Exercise 1 — level ★☆☆☆☆
Decide whether the expression
\[ 5x^2y^3 \]
is a monomial.
Answer
Yes, the expression is a monomial.
Solution
A monomial is an algebraic expression consisting of the product of a numerical coefficient and, where present, powers of variables whose exponents are non-negative integers.
In the expression
\[ 5x^2y^3 \]
the numerical factor is \(5\), whereas the literal part is
\[ x^2y^3. \]
The variable \(x\) occurs with exponent \(2\), while the variable \(y\) occurs with exponent \(3\). Both exponents are non-negative integers.
Moreover, the variables occur neither in a denominator, nor under a radical, nor as the exponent of another power. The expression therefore has precisely the structure demanded by the definition of a monomial.
Hence
\[ 5x^2y^3 \]
is a monomial with coefficient \(5\) and literal part \(x^2y^3\).
Exercise 2 — level ★☆☆☆☆
Decide whether the expression
\[ \frac{3}{x} \]
is a monomial.
Answer
No, the expression is not a monomial.
Solution
In a monomial the variables may occur only as the bases of powers whose exponents are non-negative integers.
In the expression
\[ \frac{3}{x} \]
the variable \(x\) occurs in the denominator. For \(x\neq0\), the expression may be rewritten by means of a negative exponent:
\[ \frac{3}{x}=3x^{-1}. \]
The exponent of \(x\) is thus \(-1\), a negative integer, which does not belong to the set
\[ \mathbb{N}_0=\{0,1,2,3,\dots\}. \]
The occurrence of the variable in the denominator — equivalently, the presence of a negative exponent — therefore prevents the expression from being a monomial.
Hence
\[ \frac{3}{x} \]
is not a monomial.
Exercise 3 — level ★☆☆☆☆
Reduce the monomial
\[ 2x^2\cdot(-3)y\cdot x^4y^2 \]
to standard form.
Answer
\[ -6x^6y^3 \]
Solution
A monomial is written in standard form when all the numerical factors have been multiplied together and each variable occurs exactly once, raised to its own exponent.
In the expression
\[ 2x^2\cdot(-3)y\cdot x^4y^2 \]
the numerical factors are \(2\) and \(-3\). Their product is
\[ 2\cdot(-3)=-6. \]
We now turn to the literal part. The variable \(x\) occurs in the powers
\[ x^2 \qquad\text{and}\qquad x^4. \]
By the rule for the product of powers with the same base,
\[ x^m x^n=x^{m+n}, \]
we obtain
\[ x^2x^4=x^{2+4}=x^6. \]
The variable \(y\) occurs first with no exponent written and then in the power \(y^2\). Since
\[ y=y^1, \]
it follows that
\[ y\cdot y^2=y^{1+2}=y^3. \]
Collecting the coefficient together with the literal part, we obtain
\[ 2x^2\cdot(-3)y\cdot x^4y^2 = -6x^6y^3. \]
The monomial is now written in standard form: its coefficient is \(-6\), and each variable occurs exactly once.
Exercise 4 — level ★☆☆☆☆
In the monomial
\[ -\frac{5}{2}a^3b^2 \]
identify the coefficient and the literal part.
Answer
The coefficient is
\[ -\frac{5}{2}, \]
whereas the literal part is
\[ a^3b^2. \]
Solution
In a non-zero monomial written in standard form one distinguishes two constituents: the numerical factor, sign included, is called the coefficient; the product of the powers of the variables constitutes the literal part.
Consider the monomial
\[ -\frac{5}{2}a^3b^2. \]
Its numerical factor is
\[ -\frac{5}{2}. \]
The minus sign belongs to the coefficient and must not be detached from it. The coefficient of the monomial is therefore
\[ -\frac{5}{2}. \]
The literal factors, on the other hand, are
\[ a^3 \qquad\text{and}\qquad b^2. \]
Their product makes up the literal part:
\[ a^3b^2. \]
Hence, in the monomial
\[ -\frac{5}{2}a^3b^2 \]
the coefficient is \(\displaystyle -\frac{5}{2}\), whereas the literal part is \(a^3b^2\).
Exercise 5 — level ★☆☆☆☆
Determine the degree in each variable and the total degree of the monomial
\[ -7x^3y^2z. \]
Answer
The degree in \(x\) is \(3\), the degree in \(y\) is \(2\), and the degree in \(z\) is \(1\).
The total degree of the monomial is
\[ 6. \]
Solution
The degree in a given variable is the exponent with which that variable occurs in the standard form of the monomial.
Consider
\[ -7x^3y^2z. \]
The variable \(x\) occurs in the power \(x^3\). The degree in \(x\) is therefore
\[ 3. \]
The variable \(y\) occurs in the power \(y^2\). The degree in \(y\) is accordingly
\[ 2. \]
The variable \(z\) occurs with no exponent written. In this case the exponent is implicitly equal to \(1\), since
\[ z=z^1. \]
The degree in \(z\) is thus
\[ 1. \]
The total degree of a non-zero monomial is obtained by adding the exponents of all the variables occurring in its literal part.
In the present case we have
\[ 3+2+1=6. \]
The coefficient \(-7\) has no bearing on the degree, since the degree depends solely on the exponents of the literal part.
Hence the monomial
\[ -7x^3y^2z \]
has degree \(3\) in \(x\), degree \(2\) in \(y\), degree \(1\) in \(z\) and total degree \(6\).
Exercise 6 — level ★★☆☆☆
Consider the monomials
\[ A=2x^2\cdot 3y,\qquad B=6yx^2,\qquad C=-6x^2y,\qquad D=4xy^2. \]
Determine which monomials are equal, which are like and which are opposite.
Answer
The monomials \(A\) and \(B\) are equal.
The monomials \(A\), \(B\) and \(C\) are like.
The monomials \(A\) and \(C\), as well as \(B\) and \(C\), are opposite.
The monomial \(D\) is not like the others.
Solution
In order to compare monomials properly they must first be written in standard form. In this way the coefficient and the literal part of each of them can be read off at once.
The first monomial is
\[ A=2x^2\cdot 3y. \]
Multiplying the numerical coefficients gives
\[ 2\cdot 3=6, \]
while the literal part already consists of the product \(x^2y\). Hence
\[ A=6x^2y. \]
The second monomial is
\[ B=6yx^2. \]
The order of the literal factors does not affect the product, multiplication being commutative. Arranging the variables in alphabetical order we obtain
\[ B=6x^2y. \]
The monomials \(A\) and \(B\) therefore have the same coefficient and the same literal part. They are accordingly equal.
The third monomial is
\[ C=-6x^2y. \]
Its literal part is again
\[ x^2y, \]
whereas its coefficient is \(-6\). Since \(A\), \(B\) and \(C\) have the same literal part, they are like monomials.
Two like monomials are opposite when their coefficients are opposite. The coefficients of \(A\) and \(C\) are \(6\) and \(-6\) respectively; the same holds for \(B\) and \(C\). Consequently,
\[ A+C=6x^2y-6x^2y=0 \]
and
\[ B+C=6x^2y-6x^2y=0. \]
Hence \(A\) and \(C\) are opposite, and so are \(B\) and \(C\).
Consider finally
\[ D=4xy^2. \]
Its literal part is
\[ xy^2, \]
which differs from \(x^2y\). The monomial \(D\) is therefore not like \(A\), \(B\) or \(C\).
Exercise 7 — level ★★☆☆☆
Simplify the following algebraic sum of monomials:
\[ 7x^2y-3x^2y+5xy^2-2xy^2. \]
Answer
\[ 4x^2y+3xy^2 \]
Solution
In the addition and subtraction of monomials, only like terms — that is, monomials having the same literal part — may be combined directly.
Consider the expression
\[ 7x^2y-3x^2y+5xy^2-2xy^2. \]
The monomials
\[ 7x^2y \qquad\text{and}\qquad -3x^2y \]
are like, since both have literal part
\[ x^2y. \]
Their coefficients are therefore added algebraically:
\[ 7-3=4. \]
This yields
\[ 7x^2y-3x^2y=4x^2y. \]
The monomials
\[ 5xy^2 \qquad\text{and}\qquad -2xy^2 \]
are like as well, since both have literal part
\[ xy^2. \]
Adding the coefficients gives
\[ 5-2=3, \]
and hence
\[ 5xy^2-2xy^2=3xy^2. \]
Collecting the results obtained,
\[ 7x^2y-3x^2y+5xy^2-2xy^2 = 4x^2y+3xy^2. \]
The monomials \(4x^2y\) and \(3xy^2\) are not like, since their literal parts differ. They cannot, therefore, be combined any further.
Exercise 8 — level ★★☆☆☆
Compute the product
\[ (2x^3y)(-5x^2y^4). \]
Answer
\[ -10x^5y^5 \]
Solution
The product of two monomials is always a monomial. To compute it one multiplies the numerical coefficients and adds the exponents of the powers having the same base.
Consider
\[ (2x^3y)(-5x^2y^4). \]
The numerical coefficients are \(2\) and \(-5\). Their product is
\[ 2\cdot(-5)=-10. \]
For the variable \(x\) we have
\[ x^3\cdot x^2. \]
By the rule for the product of powers with the same base,
\[ x^m x^n=x^{m+n}, \]
we obtain
\[ x^3\cdot x^2=x^{3+2}=x^5. \]
For the variable \(y\), on the other hand, we have
\[ y\cdot y^4. \]
Since \(y=y^1\),
\[ y\cdot y^4=y^{1+4}=y^5. \]
Collecting the coefficient together with the literal part, we obtain
\[ (2x^3y)(-5x^2y^4) = -10x^5y^5. \]
The result is already written in standard form.
Exercise 9 — level ★★☆☆☆
Compute the power
\[ \left(-2x^3y^2\right)^3. \]
Answer
\[ -8x^9y^6 \]
Solution
To raise a monomial to a positive integer exponent one raises the coefficient to that exponent and multiplies each exponent of the literal part by the exponent of the power.
Consider
\[ \left(-2x^3y^2\right)^3. \]
The power applies to the monomial as a whole. We may therefore write
\[ \left(-2x^3y^2\right)^3 = (-2)^3\left(x^3\right)^3\left(y^2\right)^3. \]
Let us first compute the power of the coefficient:
\[ (-2)^3=-8. \]
The result is negative because the coefficient is negative and the exponent \(3\) is odd.
For the variable \(x\) we apply the rule for a power of a power:
\[ \left(x^m\right)^n=x^{mn}. \]
Hence
\[ \left(x^3\right)^3=x^{3\cdot3}=x^9. \]
Similarly, for the variable \(y\) we have
\[ \left(y^2\right)^3=y^{2\cdot3}=y^6. \]
Collecting the coefficient together with the literal part, we obtain
\[ \left(-2x^3y^2\right)^3 = -8x^9y^6. \]
The result is already written in standard form.
Exercise 10 — level ★★☆☆☆
Decide whether the monomial
\[ 12x^5y^3 \]
is divisible by the monomial
\[ 3x^2y \]
and, if so, determine the quotient monomial.
Answer
The monomial \(12x^5y^3\) is divisible by \(3x^2y\).
The quotient monomial is
\[ 4x^3y^2. \]
Solution
A monomial \(M\) is divisible by a non-zero monomial \(N\) when there exists a monomial \(Q\) such that
\[ M=NQ. \]
In order for the quotient to be a monomial as well, the exponent of each variable in the dividend must be greater than or equal to the corresponding exponent in the divisor.
In the present case the dividend is
\[ 12x^5y^3, \]
whereas the divisor is
\[ 3x^2y. \]
Let us first compare the exponents of the variable \(x\). In the dividend the exponent is \(5\), while in the divisor it is \(2\). Since
\[ 5\geq2, \]
the required condition is met.
For the variable \(y\), the exponent in the dividend is \(3\), while in the divisor it is \(1\), since \(y=y^1\). In this case too
\[ 3\geq1. \]
The monomial \(12x^5y^3\) is therefore divisible by \(3x^2y\).
To determine the quotient one divides the coefficients:
\[ \frac{12}{3}=4, \]
and subtracts the exponents of the powers having the same base:
\[ x^{5-2}=x^3 \]
and
\[ y^{3-1}=y^2. \]
The quotient monomial is accordingly
\[ Q=4x^3y^2. \]
We check the result by multiplying the divisor by the quotient:
\[ (3x^2y)(4x^3y^2) = 12x^{2+3}y^{1+2} = 12x^5y^3. \]
The product agrees with the dividend, so the division has been carried out correctly.
Exercise 11 — level ★★☆☆☆
Decide whether the expression
\[ 3x^2y-5xy^3+7 \]
is a polynomial.
Answer
Yes, the expression is a polynomial in the variables \(x\) and \(y\).
Solution
A polynomial is a finite algebraic sum of monomials in the same variables.
Consider the expression
\[ 3x^2y-5xy^3+7. \]
To decide whether it is a polynomial, we must check that each of its terms is a monomial.
The first term is
\[ 3x^2y. \]
Its coefficient is \(3\), while the variables \(x\) and \(y\) occur with exponents \(2\) and \(1\) respectively. Both exponents are non-negative integers, so \(3x^2y\) is a monomial.
The second term is
\[ -5xy^3. \]
Its coefficient is \(-5\), while \(x\) occurs with exponent \(1\) and \(y\) with exponent \(3\). This term too is therefore a monomial.
The third term is the real number
\[ 7. \]
Every real number is a constant monomial; consequently \(7\) is a monomial as well.
The given expression is thus a finite algebraic sum of three monomials:
\[ 3x^2y,\qquad -5xy^3,\qquad 7. \]
Hence
\[ 3x^2y-5xy^3+7 \]
is a polynomial in the variables \(x\) and \(y\).
Exercise 12 — level ★★☆☆☆
Consider the polynomial
\[ P(x,y)=4x^3y-2xy^2+5x-7. \]
Identify its terms, their respective coefficients and the constant term.
Answer
The terms of the polynomial are
\[ 4x^3y,\qquad -2xy^2,\qquad 5x,\qquad -7. \]
Their respective coefficients are
\[ 4,\qquad -2,\qquad 5,\qquad -7. \]
The constant term is
\[ -7. \]
Solution
The monomials of which a polynomial is composed are called the terms of the polynomial.
Consider
\[ P(x,y)=4x^3y-2xy^2+5x-7. \]
The polynomial consists of four monomials, separated by the signs \(+\) and \(-\). Its terms are therefore
\[ 4x^3y,\qquad -2xy^2,\qquad 5x,\qquad -7. \]
The sign of each term is included in the corresponding coefficient.
In the term
\[ 4x^3y \]
the coefficient is \(4\), whereas the literal part is \(x^3y\).
In the term
\[ -2xy^2 \]
the coefficient is \(-2\), whereas the literal part is \(xy^2\).
In the term
\[ 5x \]
the coefficient is \(5\), whereas the literal part is \(x\).
The last term is
\[ -7. \]
It is a constant monomial: its coefficient is \(-7\) and its literal part is \(1\).
The term of a polynomial that contains no variables is called the constant term. In the given polynomial, accordingly, the constant term is
\[ -7. \]
In conclusion, the coefficients of the four terms are respectively
\[ 4,\qquad -2,\qquad 5,\qquad -7, \]
while the constant term is \(-7\).
Exercise 13 — level ★★☆☆☆
Reduce the polynomial
\[ 3x^2-5x+7+2x^2+4x-1. \]
Answer
\[ 5x^2-x+6 \]
Solution
A polynomial is written in reduced form when all its terms are monomials in standard form and no two of its terms are like.
Reducing a polynomial therefore amounts to identifying the like terms and combining them by adding their coefficients algebraically.
Consider the polynomial
\[ 3x^2-5x+7+2x^2+4x-1. \]
The terms
\[ 3x^2 \qquad\text{and}\qquad 2x^2 \]
are like, since both have literal part \(x^2\). Adding their coefficients gives
\[ 3+2=5, \]
and hence
\[ 3x^2+2x^2=5x^2. \]
The terms
\[ -5x \qquad\text{and}\qquad 4x \]
are like as well, since both have literal part \(x\). Adding their coefficients algebraically we get
\[ -5+4=-1. \]
Hence
\[ -5x+4x=-x. \]
Finally, the constant terms
\[ 7 \qquad\text{and}\qquad -1 \]
are like monomials, both having literal part equal to \(1\). Their sum is
\[ 7-1=6. \]
Collecting the results obtained, we have
\[ 3x^2-5x+7+2x^2+4x-1 = 5x^2-x+6. \]
The polynomial thus obtained contains no like terms and each of its terms is already written in standard form. The polynomial is therefore in reduced form.
Exercise 14 — level ★★★☆☆
Reduce the polynomial and determine its degree:
\[ P(x)=3x^4-2x^3+x^2-3x^4+5x^3-4. \]
Answer
The reduced form of the polynomial is
\[ P(x)=3x^3+x^2-4. \]
The degree of the polynomial is
\[ 3. \]
Solution
The degree of a non-zero polynomial is the largest total degree among the monomials occurring in its reduced form.
It is not enough, therefore, to look at the highest exponent appearing in the original expression: before the degree can be determined, the polynomial must be reduced by combining any like terms.
Consider
\[ P(x)=3x^4-2x^3+x^2-3x^4+5x^3-4. \]
The terms
\[ 3x^4 \qquad\text{and}\qquad -3x^4 \]
are like. Adding their coefficients gives
\[ 3-3=0. \]
The two fourth-degree terms therefore cancel:
\[ 3x^4-3x^4=0. \]
The terms
\[ -2x^3 \qquad\text{and}\qquad 5x^3 \]
are like as well. Adding their coefficients algebraically we get
\[ -2+5=3, \]
and hence
\[ -2x^3+5x^3=3x^3. \]
The terms \(x^2\) and \(-4\) have no like terms with which to be combined and therefore remain unchanged.
The reduced form of the polynomial is accordingly
\[ P(x)=3x^3+x^2-4. \]
The non-zero terms of the reduced form have degrees
\[ 3,\qquad 2,\qquad 0 \]
respectively. The largest of these is \(3\). Consequently,
\[ \deg(P)=3. \]
Although fourth-degree terms appeared in the original expression, they cancelled in the course of the reduction. The degree of the polynomial is therefore \(3\), not \(4\).
Exercise 15 — level ★★★☆☆
Reduce the following polynomials where necessary and classify them according to the number of their terms:
\[ A(x)=4x^3, \]
\[ B(x)=x^2-1, \]
\[ C(x)=x^2+3x-2x+1, \]
\[ D(x)=5x^2-5x^2. \]
Answer
\(A(x)\) is a monomial.
\(B(x)\) is a binomial.
\(C(x)\) reduces to
\[ x^2+x+1 \]
and is therefore a trinomial.
\(D(x)\) reduces to the zero polynomial and is not classified according to the number of its terms.
Solution
In order to classify a polynomial according to the number of its terms one must consider its reduced form.
A non-zero polynomial consisting of a single term is a monomial; if it has two terms it is a binomial; if it has three it is a trinomial.
Consider first
\[ A(x)=4x^3. \]
The polynomial is already written in reduced form and consists of a single non-zero term:
\[ 4x^3. \]
Hence \(A(x)\) is a monomial.
We now turn to the polynomial
\[ B(x)=x^2-1. \]
Its terms are
\[ x^2 \qquad\text{and}\qquad -1. \]
The two terms are not like and cannot therefore be combined. The polynomial is already in reduced form and has two terms.
Hence \(B(x)\) is a binomial.
Consider next
\[ C(x)=x^2+3x-2x+1. \]
Before classifying it we must combine the like terms. The terms
\[ 3x \qquad\text{and}\qquad -2x \]
have the same literal part \(x\). Adding their coefficients gives
\[ 3-2=1, \]
and hence
\[ 3x-2x=x. \]
The reduced form of \(C(x)\) is accordingly
\[ C(x)=x^2+x+1. \]
Its terms are
\[ x^2,\qquad x,\qquad 1. \]
Since the polynomial has three terms, \(C(x)\) is a trinomial.
Consider finally
\[ D(x)=5x^2-5x^2. \]
The two terms are opposite and their sum is zero:
\[ 5x^2-5x^2=0. \]
Hence
\[ D(x)=0. \]
The result is the zero polynomial. In its reduced form it has no non-zero terms and thus constitutes a special case: it is not classified as a monomial, a binomial or a trinomial according to the number of its terms.
Exercise 16 — level ★★★☆☆
Consider the polynomials
\[ A(x)=x^4-3x^3+2x^2+x-5, \]
\[ B(x)=2x^4+3x-1, \]
\[ C(x)=-4+x-2x^3+3x^4. \]
Determine, for each polynomial, whether it is ordered, complete and monic.
Answer
\(A(x)\) is ordered in decreasing powers of \(x\), complete and monic.
\(B(x)\) is ordered in decreasing powers of \(x\), incomplete and not monic.
\(C(x)\) is ordered in increasing powers of \(x\), incomplete and not monic.
Solution
A polynomial in one variable is said to be ordered when its terms are arranged in increasing or decreasing powers of the variable.
A polynomial of degree \(n\) is said to be complete when it contains, with non-zero coefficients, every power of the variable from \(n\) down to \(0\).
Finally, a non-zero polynomial in one variable is said to be monic when its leading coefficient is equal to \(1\).
Consider first
\[ A(x)=x^4-3x^3+2x^2+x-5. \]
The exponents of \(x\) occur in the order
\[ 4,\qquad 3,\qquad 2,\qquad 1,\qquad 0. \]
They are arranged in decreasing order. The polynomial is therefore ordered in decreasing powers of \(x\).
The degree of \(A(x)\) is \(4\). Every power from \(4\) down to \(0\) is present:
\[ x^4,\qquad x^3,\qquad x^2,\qquad x,\qquad x^0. \]
All the corresponding coefficients are non-zero. Hence \(A(x)\) is complete.
The leading term is
\[ x^4, \]
whose coefficient is \(1\). Consequently \(A(x)\) is monic as well.
Consider now
\[ B(x)=2x^4+3x-1. \]
The exponents of the powers present are
\[ 4,\qquad 1,\qquad 0. \]
They are arranged in decreasing order, so the polynomial is ordered in decreasing powers of \(x\).
The degree of \(B(x)\) is \(4\), but no terms in \(x^3\) or in \(x^2\) occur. The coefficients of those powers are implicitly equal to zero.
Hence \(B(x)\) is incomplete.
The leading term is
\[ 2x^4, \]
and the leading coefficient is \(2\), not \(1\). Consequently \(B(x)\) is not monic.
Consider finally
\[ C(x)=-4+x-2x^3+3x^4. \]
The exponents of the powers present are
\[ 0,\qquad 1,\qquad 3,\qquad 4. \]
They are arranged in increasing order. Hence \(C(x)\) is ordered in increasing powers of \(x\).
The degree of the polynomial is \(4\), but no term in \(x^2\) occurs. The coefficient of \(x^2\) is therefore implicitly equal to \(0\).
Consequently \(C(x)\) is incomplete.
To decide whether the polynomial is monic we must identify the coefficient of the term of highest degree. The leading term is
\[ 3x^4, \]
and the leading coefficient is \(3\). Hence \(C(x)\) is not monic.
Exercise 17 — level ★★★☆☆
Determine which of the following polynomials are homogeneous and, if so, find their degree:
\[ A(x,y)=3x^3-2x^2y+5xy^2-y^3, \]
\[ B(x,y)=x^2+xy+y, \]
\[ C(a,b,c)=2a^2b-3abc+4b^2c. \]
Answer
\(A(x,y)\) is a homogeneous polynomial of degree \(3\).
\(B(x,y)\) is not a homogeneous polynomial.
\(C(a,b,c)\) is a homogeneous polynomial of degree \(3\).
Solution
A non-zero polynomial in several variables is said to be homogeneous when all its terms have the same total degree.
To decide whether a polynomial is homogeneous we must therefore compute the total degree of each of its terms. The total degree of a monomial is obtained by adding the exponents of all the variables occurring in its literal part.
Consider first
\[ A(x,y)=3x^3-2x^2y+5xy^2-y^3. \]
The first term is
\[ 3x^3. \]
The variable \(x\) occurs with exponent \(3\), whereas \(y\) does not occur and thus has exponent \(0\). The total degree of the term is
\[ 3+0=3. \]
The second term is
\[ -2x^2y. \]
The exponents of \(x\) and \(y\) are \(2\) and \(1\) respectively. The total degree is therefore
\[ 2+1=3. \]
The third term is
\[ 5xy^2. \]
In this case the exponents are \(1\) and \(2\), so the total degree is
\[ 1+2=3. \]
The last term is
\[ -y^3. \]
The variable \(x\) does not occur and thus has exponent \(0\), while \(y\) has exponent \(3\). The total degree is accordingly
\[ 0+3=3. \]
Every term of \(A(x,y)\) has total degree \(3\). Hence \(A(x,y)\) is a homogeneous polynomial of degree \(3\).
Consider now
\[ B(x,y)=x^2+xy+y. \]
The term \(x^2\) has total degree
\[ 2. \]
The term \(xy\) has total degree
\[ 1+1=2. \]
The term \(y\), on the other hand, has total degree
\[ 1. \]
The terms do not all have the same total degree. Consequently \(B(x,y)\) is not a homogeneous polynomial.
Consider finally
\[ C(a,b,c)=2a^2b-3abc+4b^2c. \]
The term
\[ 2a^2b \]
has total degree
\[ 2+1+0=3. \]
The term
\[ -3abc \]
has total degree
\[ 1+1+1=3. \]
The term
\[ 4b^2c \]
has total degree
\[ 0+2+1=3. \]
In this case too all the terms have the same total degree. Hence \(C(a,b,c)\) is a homogeneous polynomial of degree \(3\).
Exercise 18 — level ★★★☆☆
Consider the polynomials
\[ P(x)=3x^3-2x^2+5x-1 \]
and
\[ Q(x)=-x^3+4x^2-3x+6. \]
Compute
\[ P(x)+Q(x) \]
and
\[ P(x)-Q(x). \]
Answer
The sum is
\[ P(x)+Q(x)=2x^3+2x^2+2x+5. \]
The difference is
\[ P(x)-Q(x)=4x^3-6x^2+8x-7. \]
Solution
To add or subtract two polynomials one combines the like terms, that is, the terms containing the same power of the variable.
Consider first the sum
\[ P(x)+Q(x). \]
Substituting the expressions of the two polynomials, we obtain
\[ P(x)+Q(x) = (3x^3-2x^2+5x-1)+(-x^3+4x^2-3x+6). \]
The brackets may be removed without altering the signs of the terms, since the two polynomials are separated by the sign \(+\):
\[ P(x)+Q(x) = 3x^3-2x^2+5x-1-x^3+4x^2-3x+6. \]
We now combine the like terms.
The third-degree terms are
\[ 3x^3 \qquad\text{and}\qquad -x^3. \]
Adding their coefficients gives
\[ 3-1=2, \]
and hence
\[ 3x^3-x^3=2x^3. \]
The second-degree terms are
\[ -2x^2 \qquad\text{and}\qquad 4x^2. \]
Adding their coefficients gives
\[ -2+4=2, \]
so that
\[ -2x^2+4x^2=2x^2. \]
The first-degree terms are
\[ 5x \qquad\text{and}\qquad -3x. \]
Their sum is
\[ 5x-3x=2x. \]
Finally, the constant terms are
\[ -1 \qquad\text{and}\qquad 6. \]
We have
\[ -1+6=5. \]
Hence
\[ P(x)+Q(x)=2x^3+2x^2+2x+5. \]
We now compute the difference
\[ P(x)-Q(x). \]
Substituting the expressions of the two polynomials, we obtain
\[ P(x)-Q(x) = (3x^3-2x^2+5x-1)-(-x^3+4x^2-3x+6). \]
To subtract a polynomial is to add its opposite. The minus sign placed before the second pair of brackets therefore changes the sign of every term of \(Q(x)\):
\[ -Q(x)=x^3-4x^2+3x-6. \]
Consequently,
\[ P(x)-Q(x) = 3x^3-2x^2+5x-1+x^3-4x^2+3x-6. \]
We again combine the like terms.
For the third-degree terms we have
\[ 3x^3+x^3=4x^3. \]
For the second-degree terms,
\[ -2x^2-4x^2=-6x^2. \]
For the first-degree terms,
\[ 5x+3x=8x. \]
Finally, for the constant terms,
\[ -1-6=-7. \]
Hence
\[ P(x)-Q(x)=4x^3-6x^2+8x-7. \]
Neither result contains like terms, so the polynomials obtained are already written in reduced form.
Exercise 19 — level ★★★★☆
Consider the polynomials
\[ P(x)=2x-3 \]
and
\[ Q(x)=x^2+4x+1. \]
Compute
\[ P(x)Q(x) \]
and
\[ P(x)^2. \]
Answer
The product of the two polynomials is
\[ P(x)Q(x)=2x^3+5x^2-10x-3. \]
The square of \(P(x)\) is
\[ P(x)^2=4x^2-12x+9. \]
Solution
The product of two polynomials is computed by applying the distributive law of multiplication over addition: each term of the first polynomial must be multiplied by each term of the second.
Consider first
\[ P(x)Q(x)=(2x-3)(x^2+4x+1). \]
The first polynomial consists of the terms \(2x\) and \(-3\). We multiply each of them by the whole polynomial \(Q(x)\):
\[ (2x-3)(x^2+4x+1) = 2x(x^2+4x+1)-3(x^2+4x+1). \]
Let us expand the first product:
\[ 2x(x^2+4x+1) = 2x\cdot x^2+2x\cdot4x+2x\cdot1. \]
Computing each term, we obtain
\[ 2x\cdot x^2=2x^3, \]
\[ 2x\cdot4x=8x^2 \]
and
\[ 2x\cdot1=2x. \]
Hence
\[ 2x(x^2+4x+1)=2x^3+8x^2+2x. \]
We now expand the second product:
\[ -3(x^2+4x+1) = -3\cdot x^2-3\cdot4x-3\cdot1. \]
This gives
\[ -3(x^2+4x+1)=-3x^2-12x-3. \]
Collecting the two expansions, we have
\[ P(x)Q(x) = 2x^3+8x^2+2x-3x^2-12x-3. \]
The terms
\[ 8x^2 \qquad\text{and}\qquad -3x^2 \]
are like. Adding their coefficients gives
\[ 8-3=5, \]
and hence
\[ 8x^2-3x^2=5x^2. \]
The terms
\[ 2x \qquad\text{and}\qquad -12x \]
are like as well. Their sum is
\[ 2x-12x=-10x. \]
The reduced product is accordingly
\[ P(x)Q(x)=2x^3+5x^2-10x-3. \]
We now compute the square of \(P(x)\):
\[ P(x)^2=(2x-3)^2. \]
To square a polynomial is to multiply it by itself:
\[ P(x)^2=(2x-3)(2x-3). \]
We apply the distributive law once more:
\[ (2x-3)(2x-3) = 2x(2x-3)-3(2x-3). \]
Expanding the first product we obtain
\[ 2x(2x-3)=2x\cdot2x-2x\cdot3=4x^2-6x. \]
Expanding the second product we obtain
\[ -3(2x-3)=-6x+9. \]
Collecting the results,
\[ P(x)^2=4x^2-6x-6x+9. \]
The terms \(-6x\) and \(-6x\) are like and their sum is
\[ -6x-6x=-12x. \]
Hence
\[ P(x)^2=4x^2-12x+9. \]
In both computations the result is again a polynomial, as is guaranteed by the closure of the set of polynomials under multiplication.
Exercise 20 — level ★★★★☆
Consider the polynomial
\[ P(x,y)=3x^2y-2xy^2+5. \]
Evaluate it at
\[ x=2 \qquad\text{and}\qquad y=-1. \]
Answer
\[ P(2,-1)=-11 \]
Solution
To evaluate a polynomial — that is, to compute its numerical value — one substitutes the given value for each variable and carries out the indicated operations.
The polynomial is
\[ P(x,y)=3x^2y-2xy^2+5. \]
We must substitute
\[ x=2 \qquad\text{and}\qquad y=-1. \]
The notation
\[ P(2,-1) \]
therefore denotes the number obtained by substituting \(2\) for every occurrence of \(x\) and \(-1\) for every occurrence of \(y\).
Carrying out the substitutions, we obtain
\[ P(2,-1) = 3\cdot 2^2\cdot(-1) - 2\cdot2\cdot(-1)^2 + 5. \]
The negative value assigned to \(y\) has been enclosed in brackets. This step is essential, above all when the number is to be raised to a power.
Let us first compute the powers:
\[ 2^2=4 \]
and
\[ (-1)^2=1. \]
The second power is positive because the product of two negative numbers is positive:
\[ (-1)^2=(-1)(-1)=1. \]
Substituting the values of the powers, we obtain
\[ P(2,-1) = 3\cdot4\cdot(-1) - 2\cdot2\cdot1 + 5. \]
We now compute the first product:
\[ 3\cdot4\cdot(-1)=12\cdot(-1)=-12. \]
The second product is
\[ 2\cdot2\cdot1=4. \]
Hence
\[ P(2,-1)=-12-4+5. \]
Carrying out the algebraic addition at last,
\[ -12-4=-16 \]
and
\[ -16+5=-11. \]
The required numerical value is therefore
\[ P(2,-1)=-11. \]