Solve polynomial equations up to degree four

Ontario Grade 12 Mathematics

Polynomial and Rational Functions

Choose a useful form, factor carefully, and check every solution.

A polynomial equation asks for the value or values of a variable that make a polynomial equal to zero. The degree is the greatest exponent of the variable with a nonzero coefficient. For example,

x^4-5x^2+4=0

has degree four. A polynomial of degree four can have up to four real solutions, but it may have fewer. In this lesson, you will use familiar factoring skills and the quadratic formula to solve equations up to degree four. The main idea is to rewrite an equation into a form that reveals its solutions.

What you will learn

Recognize polynomial equations of degree two, three, and four.
Use factoring, the zero-product property, and the quadratic formula to find solutions.
Choose a suitable method for a polynomial equation and check the resulting values.

1. Review: solutions, factors, and the zero-product property

A solution is a value that makes an equation true. To check a proposed solution, substitute it into the original equation. If both sides are equal, the value is a solution.

A factor is an expression multiplied by another expression. For instance,

x^2-9

can be written as

(x-3)(x+3)

. Factoring reverses expansion: it rewrites a sum or difference as a product.

The zero-product property says that if a product is zero, at least one of its factors must be zero. This lets us solve a factored equation by setting each factor equal to zero. The property applies only after one side of the equation is zero.

For a quadratic equation that does not factor easily, the quadratic formula can be used. In

ax^2+bx+c=0

,

a

,

b

, and

c

are the coefficients, and

a

must not be zero.

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

First move all terms to one side so the other side is zero.
If the left side is a product, set each factor equal to zero.
For a quadratic that is difficult to factor, use the quadratic formula.

2. Choose a structure before choosing a method

Start by arranging the equation so one side is zero. Then look for a common factor, familiar factoring pattern, or groups of terms that can be factored. Common patterns include a difference of squares, such as

u^2-v^2=(u-v)(u+v)

, and a perfect-square trinomial, such as

u^2+2uv+v^2=(u+v)^2

.

A degree-four equation may be quadratic in a repeated expression. For example, an equation involving

x^4

and

x^2

can sometimes be handled by setting

u=x^2

. This temporary substitution turns the equation into a quadratic in

u

. Solve for

u

, then return to

x

and solve

x^2=u

. Do not stop at the values of

u

; the original question asks for values of

x

.

Another useful structure is a factor that is already visible. If factoring produces a linear factor and a quadratic factor, solve the linear equation directly and solve the quadratic by factoring or by the quadratic formula. This works for cubics and quartics as well as quadratics.

Some equations do not factor readily by inspection. When a polynomial has degree three or four, a graph or a numerical method can help estimate real solutions, but an estimate should not be presented as an exact value. In this lesson, the worked equation is designed to factor exactly.

x^4+px^2+q=0,\quad u=x^2

Look for structure before expanding or applying a formula.
A substitution such as $u=x^2$ is temporary; convert the final results back to $x$ .
For each factor, solve its own equation and include every real solution.

3. A reliable solving routine

Use a consistent routine. First, write the equation with zero on one side. Second, factor as much as possible. Third, apply the zero-product property. If a factor is a quadratic that will not factor simply, use the quadratic formula. Finally, check each answer in the original equation.

Keep track of repeated factors. For example,

(x-2)^2=0

gives the solution

x=2

. It is one distinct value, even though the factor appears twice. In a list of solutions, include the value once unless the question asks about multiplicity.

A solution may be rejected by a sign error, an incomplete factorization, or a failure to return from a substitution. Checking in the original equation is a direct way to catch these problems. If a value is only an estimate from a graph or numerical method, state that it is approximate.

(A)(B)=0 \Longrightarrow A=0 or B=0

Do not apply the zero-product property to a sum that has not been factored.
Solve every factor equation.
Substitute final values into the original equation, not just a transformed one.

4. Apply the method to a quartic equation

Consider a quartic that contains both fourth-power and second-power terms. Its form suggests using

u=x^2

. The substitution is helpful because

x^4=(x^2)^2=u^2

, so the original equation becomes a quadratic. After solving that quadratic, each possible value of

u

must be translated back into an equation involving

x

.

A value of

u

that is negative cannot equal

x^2

for a real number

x

, because the square of a real number is nonnegative. A positive value of

u

leads to two real values of

x

, one positive and one negative. This is why a quartic can produce more solutions than the quadratic in

u

.

The worked example shows the full process, including the check. The check uses the original quartic, so it confirms that the final values solve the equation that was asked.

x^2=u\;\Longrightarrow\;x=\pm\sqrt{u}\quad(u>0)

Convert the quartic to a quadratic in a temporary variable.
Solve the resulting quadratic, then solve each equation of the form $x^2=u$ .
Check all final values in the original equation.

From a quartic in $x$ to a quadratic in $u$

Stage	Expression or result
Original equation	$x^4-5x^2+4=0$
Substitution	$u=x^2$
Quadratic in $u$	$u^2-5u+4=0$
Values of $u$	$u=1$ or $u=4$
Return to $x$	$x=-2,-1,1,2$

Worked example

Solve a quartic by substitution

Solve

x^4-5x^2+4=0

over the real numbers.

Identify the structure
The equation has $x^4$ and $x^2$ , so set $u=x^2$ . Then $x^4=u^2$ , and the equation becomes a quadratic in $u$ .
$u^2-5u+4=0$
Factor the quadratic
The numbers $-1$ and $-4$ multiply to $4$ and add to $-5$ . They give the two factors. Set each factor equal to zero using the zero-product property. (u-1)(u-4)=0 \Longrightarrow u=1 or u=4
Return to the original variable
Since $u=x^2$ , solve $x^2=1$ and $x^2=4$ . Each positive square value gives a positive and a negative real solution.
$x=\pm1,\;\pm2$
Check the values
For $x=1$ or $x=-1$ , the fourth power and second power are both $1$ , so the expression is zero. For $x=2$ or $x=-2$ , the powers are $16$ and $4$ , giving $16-20+4=0$ .
$1-5+4=0,\qquad16-5(4)+4=0$

Answer: The real solutions are

x=-2,-1,1,2

.

Check: Substitution into the original quartic gives zero for all four values.

Common mistakes and how to avoid them

Setting each term of a sum equal to zero before factoring.

Correction: The zero-product property applies to a product. Factor the polynomial first, then set each factor equal to zero.

Treating

u

as the final variable after using

u=x^2

.

Correction: Replace each solution for

u

with its corresponding equation in

x

. The requested solutions are values of the original variable.

Giving only the positive square root when solving

x^2=4

.

Correction: Both

2

and

-2

square to

4

, so include both real solutions.

Checking a value only in the transformed equation.

Correction: Substitute each final value into the original equation to verify that it answers the original question.

Lesson summary

Arrange a polynomial equation with zero on one side.
Factor when possible and use the zero-product property.
Use the quadratic formula when a quadratic factor does not factor readily.
For an equation in $x^4$ and $x^2$ , consider the substitution $u=x^2$ .
Return to the original variable and check every real solution.

Check your understanding

Question 1

Solve

x^4-13x^2+36=0

over the real numbers.

$x=-3,-2,2,3$
$x=2,3$
$x=-6,-1,1,6$
$x=-3,3$

Show answer and explanation

x=-3,-2,2,3

Set

u=x^2

. Then

u^2-13u+36=(u-4)(u-9)=0

, so

u=4

or

u=9

. Returning to

x

gives

x=\pm2

or

x=\pm3

.

Question 2

Which equation should be solved after substituting

u=x^2

into

x^4+3x^2-10=0

?

$u^2+3u-10=0$
$u^2+3u^2-10=0$
$u+3u-10=0$
$u^4+3u^2-10=0$

Show answer and explanation

u^2+3u-10=0

Because

x^4=(x^2)^2=u^2

and

x^2=u

, the equation becomes

u^2+3u-10=0

.

Key terms

Polynomial equation: An equation in which a polynomial expression is set equal to another expression, often zero.
Degree: The greatest exponent of the variable that has a nonzero coefficient.
Factor: An expression that is multiplied by another expression.
Zero-product property: If a product equals zero, at least one factor must equal zero.
Substitution: Replacing an expression with a temporary variable to make an equation easier to solve.

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About this lesson and its review

Published by DoAssignment. This reviewed lesson follows Ontario Grade 12 Mathematics (MHF4U), expectation C3.4. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

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