Loading [MathJax]/jax/output/HTML-CSS/jax.js
Skip to main content
Library homepage
 

Text Color

Text Size

 

Margin Size

 

Font Type

Enable Dyslexic Font
Mathematics LibreTexts

8.3: Factoring and Finding Polynomial Solutions (Zeroes)

( \newcommand{\kernel}{\mathrm{null}\,}\)

There are several ways to find solutions of polynomials which are trinomials of the form ax2+bx+c=0. These solutions are also called the real zeroes of polynomials.

  1. Trial and Check Factoring Method: With this method, the goal is to create two binomials that when multiplied together, result in the given trinomial. This method can be very difficult when the given trinomial has large values of a and c. Once the factoring is complete find all real zeroes by using the zero factor property and setting each factor equal to 0 and solve for x.
  2. Factor by Grouping Factoring Method: With this method, the goal is to create four terms by splitting the middle term into two terms, whose coefficients have a product of ac and have a sum of b. The order of the center terms does not matter. Once the four terms are created, pair the first two terms with parentheses, pair the second two terms with parentheses, and factor out the GCF from both pairs. The resultant repeated binomial is one factor, and the GCF factors combine to make the second binomial. This is the easiest method to use on any factorable trinomial of the form ax2+bx+c, but may have a bit of a learning curve. Once the factoring is complete find all real zeroes by using the zero factor property and setting each factor equal to 0 and solve for x.
  3. The Quadratic Formula: The Quadratic Formula can be used to find the real zeroes of a factorable trinomial. Please see the Table of Contents to find the section explaining how to use the Quadratic Formula.
Example 8.3.1

Factor the expressions using any of the methods discussed in this section (these example problems will demonstrate the Factor by Grouping method):

  1. 4x23x10
  2. 8x22x3
  3. 12x14x3+22x2
  4. (x2+1)2(2)+(2x)2(x2+1)(2x)(x2+1)4
  5. (2x+1)12(x+2)(2x+1)122x+1
Solution
  1. 4x23x10Example problem4x23x10Product ac is 4(10)=40, Sum is b=3. To use factor by grouping, two middle terms are needed that multiply to a product of 40 and add to a sum of 3.8 and 5 are good candidates; Since the product must be negative, one of these values must be negative.8 and positive 5 will work, because their product is 40 and their sum is 3.4x28x+5x10Four terms, the sum of the two middle terms is the original middle term, 3x(4x28x)+(5x10)Create pairs of terms4x(x2)+5(x2)Factor out the GCF from each pair- a repeated binomial factor is always present(4x+5)(x2)Solution. Be sure to check by FOIL.
  1. 8x22x3Example problem8x22x3Product ac is 8(3)=24, Sum is b=2. To use factor by grouping, two middle terms are needed that multiply to a product of 24 and add to a sum of 2.6 and 4 are good candidates; Since the product must be negative, one of these values must be negative.6 and positive 4 will work, because their product is 24 and their sum is 2.8x2+4x6x3Four terms, the sum of the two middle terms is the original middle term, 2x.The order of the two middle terms does not matter. (8x2+4x)+(6x3)Create pairs of terms. Notice the addition in between the parentheses;the third of the four terms was negative here, so the sign stays with the term.4x(2x+1)+(3)(2x+1)Factor out the GCF from each pair- a repeated binomial factor is always present(4x3)(2x+1)Solution. Be sure to check by FOIL.
  1. 12x14x3+22x2Example problem14x3+22x2+12xReorder the terms in decreasing order of variable degree.2x(7x2+11x+6)Factor out the GCF so a trinomial results that can be factored using factor by grouping.The GCF of 2x will be included in the final answer, so do not forget about it.7x2+11x+6Product ac is 76=42, Sum is b=11. To use factor by grouping, two middle terms are needed that multiply to a product of 42 and add to a sum of 11.There are no numbers that fulfill both of these requirements,which means that the trinomial is not factorable into integer factors.7x2+11x+6To find the factors and the zeroes of the polynomial, use the Quadratic Formula.Let a=7b=11c=6x=11±1124(7)(6)2(7)Quadratic Formulax=11±121+16814Simplifyx=11±28914Divide out 1 from all termsx=11±28914=2,x=1128914=37Exact answers for the zeroes in radical form, followed by real number form.(x2),(x+37)Factors. Take care to insert the correct ± in the factors.Find the solutions, and then reverse-engineer to figure out the factor that will produce that solution.The first solution from the Quadratic Formula was x=2.A factor of (x2) when set equal to 0 will produce the solution of x=2.The second solution from the Quadratic Formula was x=37.A factor of (x+37) will produce the solution of x=37.2x(x2)(x+37)Polynomial factors, including the original GCF that was factored out at the beginning of this problem.
  1. (x2+1)2(2)+(2x)2(x2+1)(2x)(x2+1)4Example problem2(x2+1)[(x2+1)(1)+(x)2(2x)](x2+1)(x2+1)3Factor out the GCF from the numerator.2(x2+1)[(x2+1)(1)+(x)2(2x)](x2+1)(x2+1)3Remove common factors.2[(x2+1)(1)+(x)2(2x)](x2+1)3Remove common factors.2[x21+4x2](x2+1)3Simplify2(3x21)(x2+1)3Final answer.
  1. (2x+1)12(x+2)(2x+1)122x+1Example problem(2x+1)12(x+2)(2x+1)122x+1Write the expression with a positive exponent (move it to the denominator).(2x+1)12(2x+1)12(x+2)(2x+1)122x+1Write the numerator with a common denominator.2x+1x2(2x+1)12(2x+1)Simplified.2x+1x2(2x+1)32Final answer.
Exercise 8.3.1

Factor using any method discussed in this section:

  1. 5x223x10
  2. 8x2+2x3
  3. 3x27x6
  4. 10x2+13x5
  5. 12x517x4+6x3
  6. (2x21)2(2)+(2x)2(2x21)(2x)(2x21)4
  7. 2(2x3)13(x1)(2x3)232x323

Quadratic Formula

Definition: Quadratic Formula

The Quadratic Formula is used to solve (or find the zeroes) of a Polynomial (Quadratic Equation) of degree 2 that is in the form ax2+bx+c=0. The Quadratic Formula is:

x=b±b24ac2a

where a, b, and c are the coefficients of the standard form of a Quadratic Equation, ax2+bx+c=0.

Example 8.3.2

For the following functions, find all zeros of f using the Quadratic Formula. Express the final answer as exact answers (in radical form) and also as decimals, rounded to the thousandths place.

  1. f(x)=2x2+4x1
  2. f(x)=x33x24x
Solution
  1. Set f(x)=0:2x2+4x1=0. This function is written in the form ax2+bx+c=0, with a=2, b=4 and c=1.

Replacing a, b and c in the Quadratic Formula with these values:

x=4±424(2)(1)2(2)Quadratic Formulax=4±(168)4Simplifyx=4±84Simplifyx=4±224Simplify the radicalx=2±22Exact answers in radical formx=222,x=2+22Exact answers written as two rootsx=0.293 and x=1.707Approximation answers rounded to the thousandths place

  1. The function f(x)=x33x24x is a cubic function. Factor out x from all three terms before using the Quadratic Equation on the trinomial factor: x(x23x4)=0, with a=1, b=3 and c=4.

Don’t forget that the x that was factored out is a root, namely x=0.

Replacing a, b and c in the Quadratic Formula with these values:

x=3±(3)24(1)(4)2(1)Quadratic Formulax=3±(16+9)2Simplifyx=3±252Simplifyx=3±52Simplify furtherx=352,x=22,x=1Second root (first root is x=0)x=3+52,x=82,x=4Third root

There are three solutions, or roots of the cubic function f(x)=x33x24x:x=0, x=1 and x=4.

Exercise 8.3.2

For the following functions, find all zeros of f using the Quadratic Formula. Express the final answer as exact answers (in radical form) and also as decimals, rounded to the thousandths place.

  1. f(t)=9t318t2+6t
  2. f(x)=x54x432x3
  3. f(x)=183x2x2
  4. f(x)=12x2+11x5
  5. f(x)=3x26x+2

Support Center

How can we help?