A progressive collection of 20 practice problems on monomials and polynomials, designed to build a rigorous understanding of the fundamental structures of elementary algebra. The exercises gradually cover identifying monomials, computing degrees, performing operations with monomials and polynomials, special products, and introductory factoring techniques.
Each solution is built step by step, with explanations that clarify not only the calculations to be carried out, but also the reasoning behind each rule being applied.
Exercise 1 — level ★☆☆☆☆
Determine whether the following expression is a monomial:
\[ 5x^2y^3 \]
Answer
Yes, it is a monomial.
Solution
A monomial is an expression obtained as the product of a numerical coefficient and powers of variables with non-negative integer exponents.
In the expression:
\[ 5x^2y^3 \]
the numerical coefficient is \(5\), and the variables are \(x\) and \(y\).
The exponents of the variables are:
\[ 2 \qquad \text{and} \qquad 3. \]
Both are non-negative integers. There are no radicals, nor negative or fractional exponents.
Therefore, the expression satisfies all the conditions required to be a monomial.
Exercise 2 — level ★☆☆☆☆
Determine whether:
\[ \frac{3}{x} \]
is a monomial.
Answer
No, it is not a monomial.
Solution
To determine whether an expression is a monomial, it is helpful to rewrite it using the properties of exponents.
Indeed, observe that:
\[ \frac{3}{x}=3x^{-1}. \]
This reveals the power:
\[ x^{-1}, \]
whose exponent is negative.
A monomial may only contain non-negative integer exponents. The presence of a negative exponent therefore violates the very definition of a monomial.
For this reason:
\[ \frac{3}{x} \]
is not a monomial.
Exercise 3 — level ★☆☆☆☆
Find the degree of the monomial:
\[ -7x^3y^2z \]
Answer
The degree of the monomial is \(6\).
Solution
The total degree of a nonzero monomial is obtained by adding the exponents of all the variables in its variable part.
In the monomial:
\[ -7x^3y^2z \]
the variables appear with the following exponents:
\[ x^3, \qquad y^2, \qquad z^1. \]
The exponent of \(z\) is implicitly \(1\), since:
\[ z=z^1. \]
Adding the exponents:
\[ 3+2+1=6. \]
The monomial therefore has total degree:
\[ 6. \]
Exercise 4 — level ★☆☆☆☆
Determine whether the following monomials are like terms:
\[ 4x^2y^3 \qquad \text{and} \qquad -9x^2y^3 \]
Answer
Yes, the two monomials are like terms.
Solution
Two monomials are called like terms when they have exactly the same variable part — that is, the same variables raised to the same exponents.
The variable part of the first monomial is:
\[ x^2y^3. \]
The second monomial has exactly the same variable part:
\[ x^2y^3. \]
Only the numerical coefficients differ, being respectively:
\[ 4 \qquad \text{and} \qquad -9. \]
Since the variable parts are identical, the two monomials are like terms.
Exercise 5 — level ★☆☆☆☆
Compute:
\[ (2x^3y)(-5x^2y^4) \]
Answer
\[ -10x^5y^5 \]
Solution
When multiplying monomials, the numerical coefficients are multiplied first, and then the exponent rules are applied to matching variables.
Multiplying the coefficients:
\[ 2\cdot(-5)=-10. \]
For the variables, the following rule is applied:
\[ x^a\cdot x^b=x^{a+b}. \]
For the variable \(x\):
\[ x^3\cdot x^2=x^{3+2}=x^5. \]
For the variable \(y\):
\[ y\cdot y^4=y^{1+4}=y^5. \]
Combining all the factors:
\[ (2x^3y)(-5x^2y^4)=-10x^5y^5. \]
Exercise 6 — level ★★☆☆☆
Compute:
\[ \frac{12x^5y^3}{3x^2y} \]
Answer
\[ 4x^3y^2 \]
Solution
When dividing monomials, the numerical coefficients are divided first, and then the following exponent rule is applied:
\[ \frac{x^a}{x^b}=x^{a-b}. \]
Starting with the coefficients:
\[ \frac{12}{3}=4. \]
For the variable \(x\):
\[ \frac{x^5}{x^2}=x^{5-2}=x^3. \]
For the variable \(y\):
\[ \frac{y^3}{y}=y^{3-1}=y^2. \]
All the resulting exponents remain non-negative, so the result is again a monomial.
Therefore:
\[ \frac{12x^5y^3}{3x^2y}=4x^3y^2. \]
Exercise 7 — level ★★☆☆☆
Simplify the following polynomial by combining like terms:
\[ 3x^2-5x+7+2x^2+4x-1 \]
Answer
\[ 5x^2-x+6 \]
Solution
Simplifying a polynomial means combining like terms — that is, terms that share the same variable part.
Observe that:
\[ 3x^2 \qquad \text{and} \qquad 2x^2 \]
are like terms, since both contain \(x^2\). Their sum is:
\[ 3x^2+2x^2=5x^2. \]
Likewise:
\[ -5x \qquad \text{and} \qquad 4x \]
are like terms. Adding the coefficients gives:
\[ -5x+4x=-x. \]
Finally, the constant terms are combined:
\[ 7-1=6. \]
The simplified polynomial is therefore:
\[ 5x^2-x+6. \]
Exercise 8 — level ★★☆☆☆
Find the degree of the polynomial:
\[ 4x^5-2x^3+x-9 \]
Answer
The degree of the polynomial is \(5\).
Solution
The degree of a nonzero polynomial equals the highest degree among its constituent monomials.
In the polynomial:
\[ 4x^5-2x^3+x-9 \]
the term of highest degree is:
\[ 4x^5, \]
since it contains the power \(x^5\).
The remaining terms have lower degree:
\[ -2x^3 \]
has degree \(3\),
\[ x \]
has degree \(1\), while:
\[ -9 \]
is a constant term and therefore has degree \(0\).
The highest degree present is thus:
\[ 5. \]
Exercise 9 — level ★★☆☆☆
Expand:
\[ (x+3)(x+5) \]
Answer
\[ x^2+8x+15 \]
Solution
To expand the product of two binomials, the distributive property of multiplication over addition is applied.
Each term of the first binomial is multiplied by each term of the second:
\[ (x+3)(x+5)=x(x+5)+3(x+5). \]
Carrying out the products:
\[ x(x+5)=x^2+5x, \]
and:
\[ 3(x+5)=3x+15. \]
Adding the results:
\[ x^2+5x+3x+15. \]
The terms:
\[ 5x \qquad \text{and} \qquad 3x \]
are like terms and can be combined:
\[ 5x+3x=8x. \]
The final result is:
\[ (x+3)(x+5)=x^2+8x+15. \]
Exercise 10 — level ★★☆☆☆
Expand:
\[ (2x-1)(x+4) \]
Answer
\[ 2x^2+7x-4 \]
Solution
Again, the distributive property is used.
Each term of the first binomial is multiplied by each term of the second:
\[ (2x-1)(x+4)=2x(x+4)-1(x+4). \]
Expanding the products:
\[ 2x(x+4)=2x^2+8x, \]
and:
\[ -1(x+4)=-x-4. \]
Adding everything together:
\[ 2x^2+8x-x-4. \]
The terms:
\[ 8x \qquad \text{and} \qquad -x \]
are like terms. Their sum is:
\[ 8x-x=7x. \]
The final result is therefore:
\[ 2x^2+7x-4. \]
Exercise 11 — level ★★☆☆☆
Expand:
\[ (x+2)^2 \]
Answer
\[ x^2+4x+4 \]
Solution
The expression:
\[ (x+2)^2 \]
represents the square of a binomial.
It is important to remember that the square of a sum is not obtained simply by squaring each term individually. The correct formula is:
\[ (a+b)^2=a^2+2ab+b^2. \]
In this case:
\[ a=x, \qquad b=2. \]
Substituting into the formula:
\[ (x+2)^2=x^2+2\cdot x\cdot2+2^2. \]
Computing the individual terms:
\[ 2\cdot x\cdot2=4x, \]
and:
\[ 2^2=4. \]
Therefore:
\[ (x+2)^2=x^2+4x+4. \]
Exercise 12 — level ★★☆☆☆
Expand:
\[ (x-5)^2 \]
Answer
\[ x^2-10x+25 \]
Solution
The expression:
\[ (x-5)^2 \]
is the square of a difference.
The following formula is therefore applied:
\[ (a-b)^2=a^2-2ab+b^2. \]
In this case:
\[ a=x, \qquad b=5. \]
Substituting:
\[ (x-5)^2=x^2-2\cdot x\cdot5+5^2. \]
Carrying out the calculations:
\[ 2\cdot x\cdot5=10x, \]
and:
\[ 5^2=25. \]
Therefore:
\[ (x-5)^2=x^2-10x+25. \]
Exercise 13 — level ★★☆☆☆
Expand:
\[ (x+3)(x-3) \]
Answer
\[ x^2-9 \]
Solution
The product:
\[ (x+3)(x-3) \]
consists of the sum and difference of the same two terms.
In such cases, the special product known as the difference of squares applies:
\[ (a+b)(a-b)=a^2-b^2. \]
Here:
\[ a=x, \qquad b=3. \]
Applying the formula directly:
\[ (x+3)(x-3)=x^2-3^2. \]
Since:
\[ 3^2=9, \]
the result is:
\[ x^2-9. \]
Exercise 14 — level ★★★☆☆
Expand:
\[ (2x-3)^2 \]
Answer
\[ 4x^2-12x+9 \]
Solution
This expression is again the square of a difference.
The following formula is applied:
\[ (a-b)^2=a^2-2ab+b^2. \]
In this case:
\[ a=2x, \qquad b=3. \]
Substituting:
\[ (2x-3)^2=(2x)^2-2(2x)(3)+3^2. \]
Computing each term in turn.
The square of the first term is:
\[ (2x)^2=4x^2. \]
The middle term is:
\[ 2(2x)(3)=12x. \]
Finally:
\[ 3^2=9. \]
Therefore:
\[ (2x-3)^2=4x^2-12x+9. \]
Exercise 15 — level ★★★☆☆
Evaluate the polynomial:
\[ P(x)=2x^2-3x+1 \]
at:
\[ x=4. \]
Answer
\[ 21 \]
Solution
Evaluating a polynomial means substituting the given value in place of the variable.
Substituting:
\[ x=4 \]
into the expression:
\[ P(x)=2x^2-3x+1. \]
This gives:
\[ P(4)=2\cdot4^2-3\cdot4+1. \]
The power is computed first:
\[ 4^2=16. \]
Therefore:
\[ P(4)=2\cdot16-12+1. \]
Carrying out the remaining operations:
\[ 2\cdot16=32, \]
and so:
\[ 32-12+1=21. \]
The required value is therefore:
\[ 21. \]
Exercise 16 — level ★★★☆☆
Find the zeros of the polynomial:
\[ x^2-7x+12 \]
Answer
\[ x=3, \qquad x=4 \]
Solution
Finding the zeros of a polynomial means finding the values of the variable that make the polynomial equal to zero.
We therefore need to solve the equation:
\[ x^2-7x+12=0. \]
We look for a factorization of the trinomial in the form:
\[ (x-a)(x-b). \]
Expanding this product gives:
\[ x^2-(a+b)x+ab. \]
Comparing with:
\[ x^2-7x+12, \]
we need two numbers such that:
\[ a+b=7 \]
and simultaneously:
\[ ab=12. \]
The numbers satisfying both conditions are:
\[ 3 \qquad \text{and} \qquad 4. \]
Therefore:
\[ x^2-7x+12=(x-3)(x-4). \]
A product is zero if and only if at least one of its factors is zero. This gives:
\[ x-3=0 \qquad \text{or} \qquad x-4=0. \]
The solutions are:
\[ x=3, \qquad x=4. \]
Exercise 17 — level ★★★☆☆
Verify that:
\[ x=2 \]
is a zero of the polynomial:
\[ P(x)=x^3-4x^2+x+6. \]
Answer
Yes, \(2\) is a zero of the polynomial.
Solution
A real number is a zero of a polynomial if substituting it for the variable makes the polynomial equal to zero.
We therefore compute:
\[ P(2). \]
Substituting \(x=2\):
\[ P(2)=2^3-4\cdot2^2+2+6. \]
Computing the powers:
\[ 2^3=8, \qquad 2^2=4. \]
Therefore:
\[ P(2)=8-4\cdot4+2+6. \]
Computing the product:
\[ 4\cdot4=16. \]
This gives:
\[ P(2)=8-16+2+6. \]
Adding step by step:
\[ 8-16=-8, \]
and then:
\[ -8+2+6=0. \]
Since:
\[ P(2)=0, \]
the number \(2\) is indeed a zero of the polynomial.
Exercise 18 — level ★★★★☆
Perform the following division using Ruffini's rule:
\[ x^3-6x^2+11x-6 \]
divided by:
\[ x-1. \]
Answer
Quotient:
\[ x^2-5x+6 \]
Remainder:
\[ 0 \]
Solution
In Ruffini's rule, the value used is:
\[ r=1, \]
since the divisor is:
\[ x-1. \]
The coefficients of the polynomial are written as:
\[ 1, \qquad -6, \qquad 11, \qquad -6. \]
The Ruffini scheme is set up as follows:
\[ \begin{array}{c|cccc} 1 & 1 & -6 & 11 & -6 \\ & & 1 & -5 & 6 \\ \hline & 1 & -5 & 6 & 0 \end{array} \]
The final number obtained is:
\[ 0, \]
which is the remainder of the division.
The coefficients:
\[ 1, \qquad -5, \qquad 6 \]
form the quotient polynomial:
\[ x^2-5x+6. \]
Therefore:
\[ x^3-6x^2+11x-6=(x-1)(x^2-5x+6). \]
Exercise 19 — level ★★★★☆
Factor:
\[ x^2-9x+20 \]
Answer
\[ (x-4)(x-5) \]
Solution
The goal is to write the trinomial in the form:
\[ (x-a)(x-b). \]
Expanding the product:
\[ (x-a)(x-b)=x^2-(a+b)x+ab. \]
Comparing with:
\[ x^2-9x+20, \]
we need two numbers such that:
\[ a+b=9 \]
and:
\[ ab=20. \]
The required numbers are:
\[ 4 \qquad \text{and} \qquad 5. \]
Therefore:
\[ x^2-9x+20=(x-4)(x-5). \]
Exercise 20 — level ★★★★★
Factor completely:
\[ x^3-6x^2+11x-6. \]
Answer
\[ (x-1)(x-2)(x-3) \]
Solution
We begin by looking for any integer zeros of the polynomial.
Since the constant term is:
\[ -6, \]
any integer zeros must be among the divisors of \(6\):
\[ \pm1, \pm2, \pm3, \pm6. \]
Testing these, we find that:
\[ P(1)=0. \]
This means that:
\[ x-1 \]
is a factor of the polynomial.
Applying Ruffini's rule yields:
\[ x^3-6x^2+11x-6=(x-1)(x^2-5x+6). \]
It remains to factor the trinomial:
\[ x^2-5x+6. \]
We look for two numbers with sum \(5\) and product \(6\). These are:
\[ 2 \qquad \text{and} \qquad 3. \]
Therefore:
\[ x^2-5x+6=(x-2)(x-3). \]
The complete factorization of the polynomial is:
\[ x^3-6x^2+11x-6=(x-1)(x-2)(x-3). \]