A progressive collection of 20 practice problems on higher-degree equations, designed to help you learn how to recognize the structure of an equation before starting any calculations. Each exercise shows not only the algebraic steps, but also why a particular method can be applied.
Recall that a higher-degree equation is a polynomial equation of degree at least \(3\). In most cases, it is not solved by applying a general formula, but by trying to transform the polynomial into a product of simpler factors.
The fundamental principle is:
\[ A\cdot B=0 \quad \Longleftrightarrow \quad A=0 \ \text{or} \ B=0 \]
For this reason, the main goal will always be the same: factor the polynomial and then set each factor equal to zero.
Exercise 1 — level ★☆☆☆☆
Solve:
\[ x^3-5x^2=0 \]
Result
\[ S=\{0,5\} \]
Solution
Initial observation
The equation is of degree three, because the highest exponent of the variable is \(3\). However, we should not immediately look for complicated formulas: first, we need to examine the structure of the polynomial.
We have:
\[ x^3-5x^2=0 \]
The two terms \(x^3\) and \(-5x^2\) have a common factor. Indeed:
\[ x^3=x^2\cdot x \]
and:
\[ -5x^2=x^2\cdot(-5) \]
Factoring out the common factor
Since both terms contain \(x^2\), we can factor out \(x^2\):
\[ x^3-5x^2=x^2(x-5) \]
Therefore, the equation becomes:
\[ x^2(x-5)=0 \]
Applying the zero product property
We now have a product equal to zero. A product is equal to zero if at least one of its factors is equal to zero.
Thus:
\[ x^2=0 \]
or:
\[ x-5=0 \]
Solving the resulting equations
From the first equation:
\[ x^2=0 \]
it follows necessarily that:
\[ x=0 \]
From the second equation:
\[ x-5=0 \]
we obtain:
\[ x=5 \]
Conclusion
The solutions of the equation are:
\[ S=\{0,5\} \]
Notice that \(x=0\) comes from the factor \(x^2\). As an element of the solution set, however, it is written only once.
Exercise 2 — level ★☆☆☆☆
Solve:
\[ x^4-9x^2=0 \]
Result
\[ S=\{-3,0,3\} \]
Solution
Analyzing the structure
The equation is:
\[ x^4-9x^2=0 \]
In this case as well, the two terms have a common factor. Indeed:
\[ x^4=x^2\cdot x^2 \]
and:
\[ -9x^2=x^2\cdot(-9) \]
Therefore, we can factor out \(x^2\).
First factorization
Factoring out \(x^2\), we get:
\[ x^4-9x^2=x^2(x^2-9) \]
Hence the equation becomes:
\[ x^2(x^2-9)=0 \]
Further factorization
The factor:
\[ x^2-9 \]
has not yet been completely factored. Since \(9=3^2\), we have:
\[ x^2-9=x^2-3^2 \]
We apply the difference of squares formula:
\[ a^2-b^2=(a-b)(a+b) \]
In our case, \(a=x\) and \(b=3\), so:
\[ x^2-9=(x-3)(x+3) \]
Complete factored form
The equation becomes:
\[ x^2(x-3)(x+3)=0 \]
The polynomial is now written as a product of factors.
Setting the factors equal to zero
By the zero product property, we set each factor equal to zero:
\[ x^2=0 \]
or:
\[ x-3=0 \]
or:
\[ x+3=0 \]
Solving
From the first equation:
\[ x^2=0 \]
we get:
\[ x=0 \]
From the second:
\[ x-3=0 \]
we get:
\[ x=3 \]
From the third:
\[ x+3=0 \]
we get:
\[ x=-3 \]
Conclusion
The solutions are:
\[ S=\{-3,0,3\} \]
Exercise 3 — level ★☆☆☆☆
Solve:
\[ x^3-8=0 \]
Result
\[ S=\{2\} \]
Solution
Recognizing the structure
The equation is:
\[ x^3-8=0 \]
There is no common factor to factor out. However, we can recognize a difference of cubes, because:
\[ 8=2^3 \]
Therefore:
\[ x^3-8=x^3-2^3 \]
Difference of cubes formula
Recall the formula:
\[ a^3-b^3=(a-b)(a^2+ab+b^2) \]
In our case:
\[ a=x,\qquad b=2 \]
Hence:
\[ x^3-2^3=(x-2)(x^2+2x+4) \]
Factored equation
The equation becomes:
\[ (x-2)(x^2+2x+4)=0 \]
We can now set the individual factors equal to zero.
First factor
From the factor:
\[ x-2=0 \]
we obtain:
\[ x=2 \]
Second factor
Now consider:
\[ x^2+2x+4=0 \]
This is a quadratic equation. To determine whether it has real solutions, we compute the discriminant:
\[ \Delta=b^2-4ac \]
Here:
\[ a=1,\qquad b=2,\qquad c=4 \]
so:
\[ \Delta=2^2-4\cdot1\cdot4 \]
that is:
\[ \Delta=4-16=-12 \]
Since the discriminant is negative, the equation has no real solutions.
Conclusion
The only real solution of the original equation is:
\[ S=\{2\} \]
Exercise 4 — level ★★☆☆☆
Solve:
\[ x^4-16=0 \]
Result
\[ S=\{-2,2\} \]
Solution
First observation
The equation is:
\[ x^4-16=0 \]
Notice that \(x^4\) can be written as a square:
\[ x^4=(x^2)^2 \]
Also:
\[ 16=4^2 \]
Therefore, the polynomial is a difference of squares:
\[ x^4-16=(x^2)^2-4^2 \]
First factorization
We apply:
\[ a^2-b^2=(a-b)(a+b) \]
with:
\[ a=x^2,\qquad b=4 \]
We obtain:
\[ x^4-16=(x^2-4)(x^2+4) \]
Checking the factorization
We must not stop too early. The factor:
\[ x^2-4 \]
is still a difference of squares, because:
\[ 4=2^2 \]
Hence:
\[ x^2-4=(x-2)(x+2) \]
Final form
The equation becomes:
\[ (x-2)(x+2)(x^2+4)=0 \]
Setting the factors equal to zero
We set each factor equal to zero.
From the first factor:
\[ x-2=0 \]
we obtain:
\[ x=2 \]
From the second factor:
\[ x+2=0 \]
we obtain:
\[ x=-2 \]
The remaining equation is:
\[ x^2+4=0 \]
that is:
\[ x^2=-4 \]
This equation has no real solutions, because the square of a real number cannot be negative.
Conclusion
The real solutions are:
\[ S=\{-2,2\} \]
Exercise 5 — level ★★☆☆☆
Solve:
\[ x^4-5x^2+4=0 \]
Result
\[ S=\{-2,-1,1,2\} \]
Solution
Recognizing the biquadratic equation
The equation is:
\[ x^4-5x^2+4=0 \]
Notice that only the following terms appear:
\[ x^4,\qquad x^2,\qquad 1 \]
Odd powers of \(x\), such as \(x^3\) or \(x\), do not appear. This suggests using a substitution.
Substitution
Set:
\[ y=x^2 \]
Then:
\[ x^4=(x^2)^2=y^2 \]
Substituting into the original equation, we get:
\[ y^2-5y+4=0 \]
We have transformed the fourth-degree equation into a quadratic equation in the variable \(y\).
Solving the equation in \(y\)
We need to solve:
\[ y^2-5y+4=0 \]
We look for two numbers whose product is \(4\) and whose sum is \(-5\). These numbers are \(-1\) and \(-4\), because:
\[ (-1)(-4)=4 \]
and:
\[ -1-4=-5 \]
Therefore:
\[ y^2-5y+4=(y-1)(y-4) \]
The equation becomes:
\[ (y-1)(y-4)=0 \]
Thus:
\[ y-1=0 \]
or:
\[ y-4=0 \]
that is:
\[ y=1 \qquad \text{or} \qquad y=4 \]
Returning to the original variable
Now we must remember that:
\[ y=x^2 \]
Therefore, the two values found for \(y\) give two equations in \(x\).
From:
\[ y=1 \]
we get:
\[ x^2=1 \]
hence:
\[ x=\pm1 \]
From:
\[ y=4 \]
we get:
\[ x^2=4 \]
hence:
\[ x=\pm2 \]
Conclusion
The solutions are:
\[ x=-2,\quad x=-1,\quad x=1,\quad x=2 \]
Therefore:
\[ S=\{-2,-1,1,2\} \]
Important observation
In biquadratic equations, one must be careful when returning from the variable \(y\) to the variable \(x\). From \(x^2=4\), for example, we do not get only \(x=2\), but also \(x=-2\).
Exercise 6 — level ★★☆☆☆
Solve:
\[ x^6-9x^3=0 \]
Result
\[ S=\left\{0,\sqrt[3]{9}\right\} \]
Solution
Initial analysis
The equation is:
\[ x^6-9x^3=0 \]
The degree of the equation is \(6\), because the highest exponent of the variable is \(6\). However, its structure is quite simple: both terms contain a common power of \(x\).
Indeed:
\[ x^6=x^3\cdot x^3 \]
and:
\[ -9x^3=x^3\cdot(-9) \]
Factoring out the common factor
We can therefore factor out \(x^3\):
\[ x^6-9x^3=x^3(x^3-9) \]
The equation becomes:
\[ x^3(x^3-9)=0 \]
Zero product property
We now have a product equal to zero. Therefore, at least one of the factors must be equal to zero:
\[ x^3=0 \]
or:
\[ x^3-9=0 \]
Solving the first factor
From the first equation:
\[ x^3=0 \]
we get:
\[ x=0 \]
Indeed, the only real number whose cube is zero is \(0\).
Solving the second factor
Now consider:
\[ x^3-9=0 \]
Move \(9\) to the right-hand side:
\[ x^3=9 \]
To solve for \(x\), take the cube root:
\[ x=\sqrt[3]{9} \]
This is a real solution, because every real number has a real cube root.
Conclusion
The real solutions of the equation are:
\[ S=\left\{0,\sqrt[3]{9}\right\} \]
Exercise 7 — level ★★☆☆☆
Solve:
\[ x^4+2x^2-3=0 \]
Result
\[ S=\{-1,1\} \]
Solution
Recognizing the form
The equation is:
\[ x^4+2x^2-3=0 \]
Notice that only even powers of the variable appear:
\[ x^4,\qquad x^2,\qquad x^0 \]
This suggests treating it as a biquadratic equation, that is, as a quadratic equation in \(x^2\).
Substitution
Set:
\[ y=x^2 \]
Then:
\[ x^4=(x^2)^2=y^2 \]
Substituting into the original equation, we obtain:
\[ y^2+2y-3=0 \]
Solving the equation in \(y\)
We need to solve:
\[ y^2+2y-3=0 \]
We look for two numbers whose product is \(-3\) and whose sum is \(2\). The numbers are \(3\) and \(-1\), since:
\[ 3\cdot(-1)=-3 \]
and:
\[ 3+(-1)=2 \]
Therefore:
\[ y^2+2y-3=(y+3)(y-1) \]
The equation becomes:
\[ (y+3)(y-1)=0 \]
Thus:
\[ y+3=0 \]
or:
\[ y-1=0 \]
that is:
\[ y=-3 \qquad \text{or} \qquad y=1 \]
Returning to the variable \(x\)
Recall that:
\[ y=x^2 \]
Therefore, the value \(y=-3\) gives:
\[ x^2=-3 \]
This equation has no real solutions, because the square of a real number cannot be negative.
The value \(y=1\), instead, gives:
\[ x^2=1 \]
hence:
\[ x=\pm1 \]
Conclusion
The real solutions are therefore:
\[ S=\{-1,1\} \]
Exercise 8 — level ★★☆☆☆
Solve:
\[ x^3+3x^2-4x-12=0 \]
Result
\[ S=\{-3,-2,2\} \]
Solution
Initial observation
The equation is:
\[ x^3+3x^2-4x-12=0 \]
There is no common factor in all terms. However, the terms can be grouped two by two in a useful way.
Write:
\[ x^3+3x^2-4x-12=(x^3+3x^2)+(-4x-12) \]
Factoring by grouping
In the first group:
\[ x^3+3x^2 \]
we can factor out \(x^2\):
\[ x^3+3x^2=x^2(x+3) \]
In the second group:
\[ -4x-12 \]
we can factor out \(-4\):
\[ -4x-12=-4(x+3) \]
Hence:
\[ x^3+3x^2-4x-12=x^2(x+3)-4(x+3) \]
Factoring out the common binomial factor
The same factor \((x+3)\) now appears in both terms:
\[ x^2(x+3)-4(x+3) \]
We can factor out \((x+3)\):
\[ x^2(x+3)-4(x+3)=(x+3)(x^2-4) \]
Further factorization
The factor:
\[ x^2-4 \]
is a difference of squares:
\[ x^2-4=(x-2)(x+2) \]
Therefore:
\[ x^3+3x^2-4x-12=(x+3)(x-2)(x+2) \]
Factored equation
The equation becomes:
\[ (x+3)(x-2)(x+2)=0 \]
Setting the factors equal to zero
Set each factor equal to zero:
\[ x+3=0 \]
or:
\[ x-2=0 \]
or:
\[ x+2=0 \]
We obtain, respectively:
\[ x=-3,\qquad x=2,\qquad x=-2 \]
Conclusion
The solutions are:
\[ S=\{-3,-2,2\} \]
Exercise 9 — level ★★★☆☆
Solve:
\[ x^3-6x^2+11x-6=0 \]
Result
\[ S=\{1,2,3\} \]
Solution
Initial observation
The equation is:
\[ x^3-6x^2+11x-6=0 \]
It cannot be factored immediately by taking out a common factor or by using a special product. In such cases, for a polynomial with integer coefficients, a natural strategy is to look for possible rational roots.
Searching for a rational root
If a polynomial with integer coefficients has an integer root, that root must divide the constant term. The constant term is \(-6\), so we test the divisors of \(6\):
\[ \pm1,\quad \pm2,\quad \pm3,\quad \pm6 \]
Let:
\[ P(x)=x^3-6x^2+11x-6 \]
Compute \(P(1)\):
\[ P(1)=1^3-6\cdot1^2+11\cdot1-6 \]
that is:
\[ P(1)=1-6+11-6=0 \]
Therefore, \(x=1\) is a root of the polynomial.
Meaning of the root found
If \(x=1\) is a root, then the polynomial is divisible by:
\[ x-1 \]
This means that we can write:
\[ x^3-6x^2+11x-6=(x-1)Q(x) \]
where \(Q(x)\) is a quadratic polynomial.
Synthetic division
Divide the polynomial by \(x-1\). The coefficients of the polynomial are:
\[ 1,\quad -6,\quad 11,\quad -6 \]
Applying synthetic division with the root \(1\), we obtain the quotient:
\[ x^2-5x+6 \]
Therefore:
\[ x^3-6x^2+11x-6=(x-1)(x^2-5x+6) \]
Factoring the quadratic trinomial
We now need to factor:
\[ x^2-5x+6 \]
We look for two numbers whose product is \(6\) and whose sum is \(-5\). The numbers are \(-2\) and \(-3\), since:
\[ (-2)(-3)=6 \]
and:
\[ -2-3=-5 \]
Therefore:
\[ x^2-5x+6=(x-2)(x-3) \]
Completely factored form
The original equation becomes:
\[ (x-1)(x-2)(x-3)=0 \]
Applying the zero product property
A product is zero if at least one of its factors is zero. Therefore:
\[ x-1=0 \]
or:
\[ x-2=0 \]
or:
\[ x-3=0 \]
Hence:
\[ x=1,\qquad x=2,\qquad x=3 \]
Conclusion
The solutions of the equation are:
\[ S=\{1,2,3\} \]
Exercise 10 — level ★★★☆☆
Solve:
\[ x^6-13x^3+36=0 \]
Result
\[ S=\left\{\sqrt[3]{4},\sqrt[3]{9}\right\} \]
Solution
Recognizing the trinomial structure
The equation is:
\[ x^6-13x^3+36=0 \]
Notice that the exponents appearing are \(6\), \(3\), and \(0\). In particular:
\[ x^6=(x^3)^2 \]
This suggests a substitution similar to the one used for quadratic equations.
Substitution
Set:
\[ y=x^3 \]
Then:
\[ x^6=(x^3)^2=y^2 \]
Substituting into the equation, we obtain:
\[ y^2-13y+36=0 \]
Solving the equation in \(y\)
Factor the trinomial:
\[ y^2-13y+36 \]
We look for two numbers whose product is \(36\) and whose sum is \(-13\). The numbers are \(-4\) and \(-9\), because:
\[ (-4)(-9)=36 \]
and:
\[ -4-9=-13 \]
Therefore:
\[ y^2-13y+36=(y-4)(y-9) \]
The equation becomes:
\[ (y-4)(y-9)=0 \]
Thus:
\[ y=4 \qquad \text{or} \qquad y=9 \]
Returning to the original variable
Since we set:
\[ y=x^3 \]
we must solve:
\[ x^3=4 \]
or:
\[ x^3=9 \]
Over the real numbers, every equation of the form \(x^3=a\) has exactly one real solution:
\[ x=\sqrt[3]{a} \]
Therefore, we get:
\[ x=\sqrt[3]{4} \]
or:
\[ x=\sqrt[3]{9} \]
Conclusion
The real solutions are:
\[ S=\left\{\sqrt[3]{4},\sqrt[3]{9}\right\} \]
Exercise 11 — level ★★★☆☆
Solve:
\[ x^5-4x^3=0 \]
Result
\[ S=\{-2,0,2\} \]
Solution
Initial analysis
The equation is:
\[ x^5-4x^3=0 \]
The highest exponent is \(5\), so this is a fifth-degree equation. However, we should not be misled by the degree: the polynomial has an evident common factor.
Indeed:
\[ x^5=x^3\cdot x^2 \]
and:
\[ -4x^3=x^3\cdot(-4) \]
Factoring
Factor out \(x^3\):
\[ x^5-4x^3=x^3(x^2-4) \]
The equation becomes:
\[ x^3(x^2-4)=0 \]
Factoring the second factor
The factor:
\[ x^2-4 \]
is a difference of squares, because \(4=2^2\). Hence:
\[ x^2-4=(x-2)(x+2) \]
Therefore:
\[ x^5-4x^3=x^3(x-2)(x+2) \]
Factored form of the equation
The equation can be rewritten as:
\[ x^3(x-2)(x+2)=0 \]
Setting the factors equal to zero
Now set each factor equal to zero:
\[ x^3=0 \]
or:
\[ x-2=0 \]
or:
\[ x+2=0 \]
From the first equation we get:
\[ x=0 \]
From the second:
\[ x=2 \]
From the third:
\[ x=-2 \]
Conclusion
The solutions are:
\[ S=\{-2,0,2\} \]
Although \(x=0\) comes from the factor \(x^3\), it is written only once in the solution set.
Exercise 12 — level ★★★☆☆
Solve:
\[ x^4-10x^2+9=0 \]
Result
\[ S=\{-3,-1,1,3\} \]
Solution
Recognizing the structure
The equation is:
\[ x^4-10x^2+9=0 \]
Only even powers of \(x\) appear: \(x^4\), \(x^2\), and the constant term. This tells us that the equation is biquadratic.
Substitution
Set:
\[ y=x^2 \]
Then:
\[ x^4=(x^2)^2=y^2 \]
Substituting, we obtain:
\[ y^2-10y+9=0 \]
Solving the equation in \(y\)
We look for two numbers whose product is \(9\) and whose sum is \(-10\). The numbers are \(-1\) and \(-9\), since:
\[ (-1)(-9)=9 \]
and:
\[ -1-9=-10 \]
Therefore:
\[ y^2-10y+9=(y-1)(y-9) \]
The equation becomes:
\[ (y-1)(y-9)=0 \]
Hence:
\[ y=1 \qquad \text{or} \qquad y=9 \]
Returning to the original variable
Since \(y=x^2\), we must solve:
\[ x^2=1 \]
or:
\[ x^2=9 \]
From the first equation:
\[ x=\pm1 \]
From the second:
\[ x=\pm3 \]
Conclusion
The solutions are:
\[ S=\{-3,-1,1,3\} \]
Exercise 13 — level ★★★☆☆
Solve:
\[ x^3+x^2-4x-4=0 \]
Result
\[ S=\{-2,-1,2\} \]
Solution
Initial observation
The equation is:
\[ x^3+x^2-4x-4=0 \]
There is no common factor shared by all four terms. In such cases, it can be useful to try factoring by grouping, grouping the terms so that the same factor appears.
Grouping the terms
Write:
\[ x^3+x^2-4x-4=(x^3+x^2)+(-4x-4) \]
In the first group, we can factor out \(x^2\):
\[ x^3+x^2=x^2(x+1) \]
In the second group, we can factor out \(-4\):
\[ -4x-4=-4(x+1) \]
Hence:
\[ x^3+x^2-4x-4=x^2(x+1)-4(x+1) \]
Factoring out the common factor
Both terms now contain the factor \((x+1)\):
\[ x^2(x+1)-4(x+1) \]
Factoring out \((x+1)\), we obtain:
\[ x^2(x+1)-4(x+1)=(x+1)(x^2-4) \]
Factoring the difference of squares
The factor:
\[ x^2-4 \]
is a difference of squares:
\[ x^2-4=(x-2)(x+2) \]
Therefore:
\[ x^3+x^2-4x-4=(x+1)(x-2)(x+2) \]
Factored equation
The equation becomes:
\[ (x+1)(x-2)(x+2)=0 \]
Setting the factors equal to zero
Set each factor equal to zero:
\[ x+1=0 \]
or:
\[ x-2=0 \]
or:
\[ x+2=0 \]
We obtain:
\[ x=-1,\qquad x=2,\qquad x=-2 \]
Conclusion
The solutions are:
\[ S=\{-2,-1,2\} \]
Exercise 14 — level ★★★★☆
Solve:
\[ x^3-3x^2-4x+12=0 \]
Result
\[ S=\{-2,2,3\} \]
Solution
Analyzing the polynomial
The equation is:
\[ x^3-3x^2-4x+12=0 \]
Here too, there is no common factor in all terms. We therefore try to group the terms in such a way that a common binomial factor appears.
Factoring by grouping
Group as follows:
\[ x^3-3x^2-4x+12=(x^3-3x^2)+(-4x+12) \]
In the first group, factor out \(x^2\):
\[ x^3-3x^2=x^2(x-3) \]
In the second group, factor out \(-4\):
\[ -4x+12=-4(x-3) \]
Therefore:
\[ x^3-3x^2-4x+12=x^2(x-3)-4(x-3) \]
Factoring out the common factor
The common factor is \((x-3)\). Factoring it out:
\[ x^2(x-3)-4(x-3)=(x-3)(x^2-4) \]
Complete factorization
Since:
\[ x^2-4=(x-2)(x+2) \]
we obtain:
\[ x^3-3x^2-4x+12=(x-3)(x-2)(x+2) \]
Solving
The equation becomes:
\[ (x-3)(x-2)(x+2)=0 \]
By the zero product property:
\[ x-3=0 \]
or:
\[ x-2=0 \]
or:
\[ x+2=0 \]
Hence:
\[ x=3,\qquad x=2,\qquad x=-2 \]
Conclusion
Writing the solutions in increasing order:
\[ S=\{-2,2,3\} \]
Exercise 15 — level ★★★★☆
Solve:
\[ x^4+x^2-6=0 \]
Result
\[ S=\{-\sqrt{2},\sqrt{2}\} \]
Solution
Recognizing the biquadratic form
The equation is:
\[ x^4+x^2-6=0 \]
Only \(x^4\), \(x^2\), and the constant term appear. We can therefore introduce a new variable:
\[ y=x^2 \]
From this substitution it follows that:
\[ x^4=(x^2)^2=y^2 \]
Equation in the new variable
Replacing \(x^2\) with \(y\) and \(x^4\) with \(y^2\), the equation becomes:
\[ y^2+y-6=0 \]
We have thus transformed a fourth-degree equation into a quadratic equation.
Factoring the trinomial
We look for two numbers whose product is \(-6\) and whose sum is \(1\). The numbers are \(3\) and \(-2\), because:
\[ 3\cdot(-2)=-6 \]
and:
\[ 3+(-2)=1 \]
Therefore:
\[ y^2+y-6=(y+3)(y-2) \]
The equation becomes:
\[ (y+3)(y-2)=0 \]
Solutions in \(y\)
By the zero product property:
\[ y+3=0 \]
or:
\[ y-2=0 \]
Hence:
\[ y=-3 \qquad \text{or} \qquad y=2 \]
Returning to the variable \(x\)
Now recall that:
\[ y=x^2 \]
The value \(y=-3\) gives:
\[ x^2=-3 \]
This equation has no real solutions, because the square of a real number is always greater than or equal to zero.
The value \(y=2\), instead, gives:
\[ x^2=2 \]
hence:
\[ x=\pm\sqrt{2} \]
Conclusion
The real solutions of the original equation are:
\[ S=\{-\sqrt{2},\sqrt{2}\} \]
Exercise 16 — level ★★★★☆
Solve:
\[ x^3-2x^2-x+2=0 \]
Result
\[ S=\{-1,1,2\} \]
Solution
Initial observation
The equation is:
\[ x^3-2x^2-x+2=0 \]
We cannot factor out a common factor from all terms. We therefore try grouping the terms:
\[ x^3-2x^2-x+2=(x^3-2x^2)+(-x+2) \]
Factoring by grouping
In the first group, factor out \(x^2\):
\[ x^3-2x^2=x^2(x-2) \]
In the second group, factor out \(-1\):
\[ -x+2=-(x-2) \]
Therefore:
\[ x^3-2x^2-x+2=x^2(x-2)-(x-2) \]
Factoring out the common binomial
Both terms contain the factor \((x-2)\). Factoring it out, we obtain:
\[ x^2(x-2)-(x-2)=(x-2)(x^2-1) \]
Factoring the difference of squares
The factor:
\[ x^2-1 \]
is a difference of squares, because:
\[ 1=1^2 \]
Hence:
\[ x^2-1=(x-1)(x+1) \]
Consequently:
\[ x^3-2x^2-x+2=(x-2)(x-1)(x+1) \]
Factored equation
The equation becomes:
\[ (x-2)(x-1)(x+1)=0 \]
Setting the factors equal to zero
Set each factor equal to zero:
\[ x-2=0 \]
or:
\[ x-1=0 \]
or:
\[ x+1=0 \]
We obtain:
\[ x=2,\qquad x=1,\qquad x=-1 \]
Conclusion
Writing the solutions in increasing order:
\[ S=\{-1,1,2\} \]
Exercise 17 — level ★★★★☆
Solve:
\[ x^4-2x^3-7x^2+8x+12=0 \]
Result
\[ S=\{-2,-1,2,3\} \]
Solution
Choosing a strategy
The polynomial:
\[ x^4-2x^3-7x^2+8x+12 \]
has no evident common factor and is not immediately recognizable as a special product. In such cases, we can look for rational roots.
Let:
\[ P(x)=x^4-2x^3-7x^2+8x+12 \]
Searching for an integer root
Since the constant term is \(12\), any integer roots must be among the divisors of \(12\):
\[ \pm1,\quad \pm2,\quad \pm3,\quad \pm4,\quad \pm6,\quad \pm12 \]
Try \(x=2\):
\[ P(2)=2^4-2\cdot2^3-7\cdot2^2+8\cdot2+12 \]
Compute each term:
\[ 2^4=16,\qquad -2\cdot2^3=-16,\qquad -7\cdot2^2=-28,\qquad 8\cdot2=16 \]
Therefore:
\[ P(2)=16-16-28+16+12=0 \]
Hence \(x=2\) is a root, and the polynomial is divisible by \(x-2\).
First Ruffini division
Dividing:
\[ x^4-2x^3-7x^2+8x+12 \]
by \(x-2\), we obtain:
\[ x^3-7x-6 \]
Therefore:
\[ P(x)=(x-2)(x^3-7x-6) \]
Factoring the cubic polynomial
We now need to factor:
\[ x^3-7x-6 \]
We look for another integer root. Try \(x=3\):
\[ 3^3-7\cdot3-6=27-21-6=0 \]
So \(x=3\) is a root of the cubic polynomial, and we can divide by \(x-3\).
Second Ruffini division
Dividing:
\[ x^3-7x-6 \]
by \(x-3\), we obtain:
\[ x^2+3x+2 \]
Thus:
\[ x^3-7x-6=(x-3)(x^2+3x+2) \]
Final factorization
Factor:
\[ x^2+3x+2 \]
We look for two numbers whose product is \(2\) and whose sum is \(3\). They are \(1\) and \(2\), so:
\[ x^2+3x+2=(x+1)(x+2) \]
The original polynomial is therefore:
\[ P(x)=(x-2)(x-3)(x+1)(x+2) \]
Solving the equation
The equation becomes:
\[ (x-2)(x-3)(x+1)(x+2)=0 \]
Hence:
\[ x=2,\qquad x=3,\qquad x=-1,\qquad x=-2 \]
Conclusion
In increasing order:
\[ S=\{-2,-1,2,3\} \]
Exercise 18 — level ★★★★★
Solve:
\[ x^6-7x^3-8=0 \]
Result
\[ S=\{-1,2\} \]
Solution
Recognizing the structure
The equation is:
\[ x^6-7x^3-8=0 \]
The exponents that appear are \(6\), \(3\), and \(0\). Since:
\[ x^6=(x^3)^2 \]
we can treat the equation as a quadratic equation in the quantity \(x^3\).
Substitution
Set:
\[ y=x^3 \]
Then:
\[ x^6=y^2 \]
The equation becomes:
\[ y^2-7y-8=0 \]
Solving the equation in \(y\)
Factor:
\[ y^2-7y-8 \]
We look for two numbers whose product is \(-8\) and whose sum is \(-7\). The numbers are \(-8\) and \(1\), since:
\[ (-8)\cdot1=-8 \]
and:
\[ -8+1=-7 \]
Therefore:
\[ y^2-7y-8=(y-8)(y+1) \]
The equation becomes:
\[ (y-8)(y+1)=0 \]
Hence:
\[ y=8 \]
or:
\[ y=-1 \]
Returning to the variable \(x\)
Since:
\[ y=x^3 \]
we obtain two equations:
\[ x^3=8 \]
or:
\[ x^3=-1 \]
From the first:
\[ x=\sqrt[3]{8}=2 \]
From the second:
\[ x=\sqrt[3]{-1}=-1 \]
Conclusion
The real solutions are:
\[ S=\{-1,2\} \]
Exercise 19 — level ★★★★★
Solve:
\[ (x-1)^2(x+2)=0 \]
Result
\[ S=\{-2,1\} \]
Solution
Equation already in factored form
The equation is:
\[ (x-1)^2(x+2)=0 \]
In this case, the polynomial is already written as a product of factors. We therefore do not need to factor it further: we can apply the zero product property directly.
Setting the factors equal to zero
A product is zero if at least one of its factors is zero. Therefore:
\[ (x-1)^2=0 \]
or:
\[ x+2=0 \]
First factor
From the first equation:
\[ (x-1)^2=0 \]
it follows that:
\[ x-1=0 \]
and therefore:
\[ x=1 \]
The square indicates that the root \(x=1\) appears twice in the factorization. We say that \(x=1\) is a double root.
Second factor
From the second equation:
\[ x+2=0 \]
we obtain:
\[ x=-2 \]
Conclusion
As a solution set, each value is written only once:
\[ S=\{-2,1\} \]
The multiplicity of the root \(1\) is important when studying the graph of the polynomial, but it does not change the solution set.
Exercise 20 — level ★★★★★
Solve:
\[ x^4-4x^3+4x^2=0 \]
Result
\[ S=\{0,2\} \]
Solution
Initial analysis
The equation is:
\[ x^4-4x^3+4x^2=0 \]
All terms contain at least the factor \(x^2\). Indeed:
\[ x^4=x^2\cdot x^2 \]
\[ -4x^3=x^2\cdot(-4x) \]
\[ 4x^2=x^2\cdot4 \]
Factoring out the common factor
Factor out \(x^2\):
\[ x^4-4x^3+4x^2=x^2(x^2-4x+4) \]
The equation becomes:
\[ x^2(x^2-4x+4)=0 \]
Recognizing the square of a binomial
Consider the trinomial:
\[ x^2-4x+4 \]
This is a perfect square trinomial, because:
\[ x^2-4x+4=x^2-2\cdot x\cdot2+2^2 \]
hence:
\[ x^2-4x+4=(x-2)^2 \]
Complete factored form
The equation becomes:
\[ x^2(x-2)^2=0 \]
Setting the factors equal to zero
Set the factors equal to zero:
\[ x^2=0 \]
or:
\[ (x-2)^2=0 \]
From the first equation:
\[ x=0 \]
From the second:
\[ x-2=0 \]
therefore:
\[ x=2 \]
Conclusion
The solutions are:
\[ S=\{0,2\} \]
Both roots have multiplicity \(2\), because they appear through the quadratic factors \(x^2\) and \((x-2)^2\).