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Factorization of Quadratic Expressions

Factor quadratics with positive coefficients

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Factoring Review

To factor means to write an expression as a product instead of a sum.  Factoring is particularly useful when solving equations set equal to zero because then logically at least one factor must be equal to zero.  In PreCalculus, you should be able to factor even when there is no obvious greatest common factor or the difference is not between two perfect squares. 

How do you use the difference of perfect squares factoring technique on polynomials that don’t contain perfect squares and why would this be useful? 

Factoring Functions

A polynomial is a sum of a finite number of terms.  Each term consists of a constant that multiplies a variable.  The variable may only be raised to a non-negative exponent.  The letters \begin{align*}a, b, c \ldots\end{align*}a,b,c in the following general polynomial expression stand for regular numbers like \begin{align*}0, 5, -\frac{1}{4}, \sqrt{2}\end{align*}0,5,14,2 and the \begin{align*}x\end{align*}x represents the variable. 

\begin{align*}ax^n+bx^{n-1}+ \ldots + fx^2+gx+h\end{align*}axn+bxn1++fx2+gx+h

You have already learned many properties of polynomials.  For example, you know the commutative property which states that terms of a polynomial can be rearranged to create an equivalent polynomial.  When two polynomials are added, subtracted or multiplied the result is always a polynomial.  This means polynomials are closed under addition, and is one of the properties that makes the factoring of polynomials possible.  Polynomials are not closed under division because dividing two polynomials could result in a variable in the denominator, which is a rational expression (not a polynomial). 

There are three methods for factoring that are essential to master.  

Greatest Common Factor Method

The first method you should always try is to factor out the greatest common factor (GCF) of the expression.

To factor the following expression, first apply the GCF method:


Many students just learning factoring may conclude that the three terms share no factors besides one.  However, the name GCF is deceiving because this expression has an infinite number of equivalent expressions many of which are more useful.  In order to find these alternative expressions you must strategically factor numbers that are neither the greatest factor nor common to all three terms. In this case, \begin{align*}-\frac{1}{2}\end{align*}12 is an excellent choice.


In order to check to see that this is an equivalent expression, you need to distribute the \begin{align*}-\frac{1}{2}\end{align*}12. When you distribute, the first coefficient matches because it just gets multiplied by 1, the second term becomes \begin{align*}\frac{7}{2}\end{align*}72 and the third term becomes -6. Note that this expression is not completely factored yet but it is simplified as much as it can be with just the GCF method. 

Factoring Into Binomials Method

The second method you should implement after factoring out a GCF is to see if you can factor the expression into the product of two binomials. This type of factoring is usually recognizable as a trinomial where \begin{align*}x^2\end{align*}x2 has a coefficient of 1. 

To continue factoring the expression from the Greatest Common Factor section, factor the following expression into the product of two binomials and a constant: 


Many students familiar with basic factoring may be initially stuck on a problem like this.  However, you should recognize that beneath the \begin{align*}4^{th}\end{align*}4th degree and the \begin{align*}-\frac{1}{2}\end{align*}12 the problem boils down to being able to factor \begin{align*}u^2-7u+12\end{align*}u27u+12 which is just \begin{align*}(u-3)(u-4)\end{align*}(u3)(u4)

Start by rewriting the problem: \begin{align*}-\frac{1}{2}(x^4-7x^2+12)\end{align*}12(x47x2+12)

Then choose a temporary substitution: Let \begin{align*}u=x^2\end{align*}u=x2

Then substitute and factor away.  Remember to substitute back at the end.

\begin{align*}-\frac{1}{2}(u^2-7u+12) &=-\frac{1}{2}(u-3)(u-4)\\ &= -\frac{1}{2}(x^2-3)(x^2-4)\end{align*}12(u27u+12)=12(u3)(u4)=12(x23)(x24)

This type of temporary substitution that enables you to see the underlying structure of an expression is very common in calculus. The expression is still not completely factored and since there are no more trinomials, you must apply the last method. 

Difference of Squares Method

The third method of basic factoring is the difference of squares.  It is recognizable as one square monomial being subtracted from another square monomial. 

To finish factoring the resulting expression from the Factoring Into Binomials section, factor the expression into four linear factors and a constant:


Many students may recognize that \begin{align*}x^2-4\end{align*}x24 immediately factors by the difference of squares method to be \begin{align*}(x-2)(x+2)\end{align*}(x2)(x+2).  This problem asks for more because sometimes the difference of squares method can be applied to expressions like \begin{align*}x^2-3\end{align*}x23 where each term is not a perfect square.  The number 3 actually is a square. 


So the fully factored expression would be:

\begin{align*}-\frac{1}{2} \left(x- \sqrt{3}\right)\left(x+ \sqrt{3}\right)(x-2)(x+2)\end{align*}12(x3)(x+3)(x2)(x+2)


Example 1

Earlier, you were asked how you use the difference of perfect squares factoring technique on polynomials that don'at contain perfect squares and why it would be useful.  One reason why it might be useful to completely factor an expression like \begin{align*}-\frac{1}{2}(x^4-7x^2+12)\end{align*}12(x47x2+12) into linear factors is if you wanted to find the roots of the function \begin{align*}f(x)=-\frac{1}{2}(x^4-7x^2+12)\end{align*}f(x)=12(x47x2+12).  The roots are \begin{align*}x=\pm \sqrt{3}, \pm 2\end{align*}x=±3,±2

You should recognize that \begin{align*}x^2-3\end{align*}x23 can still be thought of as the difference of perfect squares because the number 3 can be expressed as \begin{align*}\left(\sqrt{3}\right)^2\end{align*}(3)2.  Rewriting the number 3 to fit a factoring pattern that you already know is an example of using the basic factoring techniques at a PreCalculus level. 

Example 2

Factor the following expression into strictly linear factors if possible.  If not possible, explain why. 

\begin{align*}\frac{x^5}{3}- \frac{11x^3}{3}+6x &\\ &= \frac{1}{3}x(x^4-11x^2+18)\\ &= \frac{1}{3}x(x^2 -2)(x^2-9)\\ &= \frac{1}{3}x(x+ \sqrt2)(x- \sqrt2)(x+3)(x-3) \end{align*}x5311x33+6x=13x(x411x2+18)=13x(x22)(x29)=13x(x+2)(x2)(x+3)(x3) 
Example 3

Factor the following expression into strictly linear factors if possible.  If not possible, explain why.


For \begin{align*}-\frac{2}{7}x^4+\frac{74}{63}x^2-\frac{8}{63}\end{align*} , let \begin{align*}u=x^2\end{align*}.

\begin{align*}&= -\frac{2}{7}u^2+\frac{74}{63}u-\frac{8}{63}\\ &= -\frac{2}{7} \left(u^2-\frac{37}{9}u+\frac{4}{9}\right)\end{align*}

Factoring through fractions like this can be extremely tricky.  You must recognize that \begin{align*}-\frac{1}{9}\end{align*} and -4 sum to \begin{align*}-\frac{37}{9}\end{align*} and multiply to \begin{align*}\frac{4}{9}\end{align*}

\begin{align*}&= -\frac{2}{7} \left(u-\frac{1}{9}\right)(u-4)\\ &= -\frac{2}{7} \left(x^2-\frac{1}{9}\right)(x^2-4)\\ &= -\frac{2}{7} \left(x-\frac{1}{3}\right)\left(x+\frac{1}{3}\right)(x-2)(x+2)\end{align*}

Example 4

Factor the following expression into strictly linear factors if possible.  If not possible, explain why. 


\begin{align*}x^4+x^2-72\end{align*} \begin{align*}=(x^2-8)(x^2+9)\end{align*}

Notice that \begin{align*}(x^2-8)\end{align*} can be written as the difference of perfect squares because \begin{align*}8=\left(\sqrt{8}\right)^2=\left(2\sqrt{2}\right)^2\end{align*}.  On the other hand, \begin{align*}x^2+9\end{align*} cannot be written as the difference between squares because the \begin{align*}x^2\end{align*} and the 9 are being added not subtracted.  This polynomial cannot be factored into strictly linear factors.

\begin{align*}x^4+x^2-72\end{align*}\begin{align*}=(x-2 \sqrt{2})(x+2 \sqrt{2})(x^2+9)\end{align*}


Factor each polynomial into strictly linear factors if possible.  If not possible, explain why not.

  1. \begin{align*}x^2+5x+6\end{align*}
  2. \begin{align*}x^4+5x^2+6\end{align*}
  3. \begin{align*}x^4-16\end{align*}
  4. \begin{align*}2x^2-20\end{align*}
  5. \begin{align*}3x^2+9x+6\end{align*}
  6. \begin{align*}\frac{x^4}{2}-5x^2+\frac{9}{2}\end{align*}
  7. \begin{align*}\frac{2x^4}{3}-\frac{34x^2}{3}+\frac{32}{3}\end{align*}
  8. \begin{align*}x^2-\frac{1}{4}\end{align*}
  9. \begin{align*}x^4-\frac{37x^2}{4}+\frac{9}{4}\end{align*}
  10. \begin{align*}\frac{3}{4}x^4-\frac{87}{4}x^2+75\end{align*}
  11. \begin{align*}\frac{1}{2}x^4-\frac{29}{2}x^2+50\end{align*}
  12. \begin{align*}\frac{x^4}{2}-\frac{5x^2}{9}+\frac{1}{18}\end{align*}
  13. \begin{align*}x^4-\frac{13}{36}x^2+\frac{1}{36}\end{align*}
  14. How does the degree of a polynomial relate to the number of linear factors?
  15. If a polynomial does not have strictly linear factors, what does this imply about the type of roots that the polynomial has?

Review (Answers)

To see the Review answers, open this PDF file and look for section 2.1. 

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Greatest Common Factor The greatest common factor of two numbers is the greatest number that both of the original numbers can be divided by evenly.
linear factors Linear factors are expressions of the form ax+b where a and b are real numbers.
Polynomial A polynomial is an expression with at least one algebraic term, but which does not indicate division by a variable or contain variables with fractional exponents.
Quadratic Formula The quadratic formula states that for any quadratic equation in the form ax^2+bx+c=0, x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}.
Quadratic Polynomials A quadratic polynomial is a polynomial of the 2nd degree, in other words, a polynomial with an x^2 term.
Trinomial A trinomial is a mathematical expression with three terms.

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