<meta http-equiv="refresh" content="1; url=/nojavascript/"> Factoring Polynomials in Quadratic Form | CK-12 Foundation
Dismiss
Skip Navigation
You are reading an older version of this FlexBook® textbook: CK-12 Algebra II with Trigonometry Concepts Go to the latest version.

6.8: Factoring Polynomials in Quadratic Form

Difficulty Level: At Grade Created by: CK-12
 0  0  0

The volume of a rectangular prism is 10x^3 - 25x^2 - 15x . What are the lengths of the prism's sides?

Guidance

The last type of factorable polynomial are those that are in quadratic form. Quadratic form is when a polynomial looks like a trinomial or binomial and can be factored like a quadratic. One example is when a polynomial is in the form ax^4+bx^2+c . Another possibility is something similar to the difference of squares, a^4-b^4 . This can be factored to (a^2-b^2)(a^2+b^2) or (a-b)(a+b)(a^2+b^2) . Always keep in mind that the greatest common factors should be factored out first.

Example A

Factor 2x^4-x^2-15 .

Solution: This particular polynomial is factorable. Let’s use the method we learned in the Factoring when a \neq 1 concept. First, ac = -30 . The factors of -30 that add up to -1 are -6 and 5. Expand the middle term and then use factoring by grouping.

& 2x^4-x^2-15\\& 2x^4-6x^2+5x^2-15\\& 2x^2(x^2-3)+5(x^2-3)\\& (x^2-3)(2x^2+5)

Both of the factors are not factorable, so we are done.

Example B

Factor 81x^4-16 .

Solution: Treat this polynomial equation like a difference of squares.

& 81x^4-16\\& (9x^2-4)(9x^2+4)

Now, we can factor 9x^2-4 using the difference of squares a second time.

(3x-2)(3x+2)(9x^2+4)

9x^2+4 cannot be factored because it is a sum of squares. This will have imaginary solutions.

Example C

Find all the real-number solutions of 6x^5-51x^3-27x = 0 .

Solution: First, pull out the GCF among the three terms.

6x^5-51x^3-27x &= 0\\3x(2x^4-17x^2-9) &= 0

Factor what is inside the parenthesis like a quadratic equation. ac = -18 and the factors of -18 that add up to -17 are -18 and 1. Expand the middle term and then use factoring by grouping.

6x^5-51x^3-27x &= 0\\3x(2x^4-17x^2-9) &= 0\\3x(2x^4-18x^2+x^2-9) &= 0\\3x[2x^2(x^2-9)+1(x^2-9)] &= 0\\3x(x^2-9)(2x^2+1) &= 0

Factor x^2-9 further and solve for x where possible. 2x^2+1 is not factorable.

3x(x^2-9)(2x^2+1) &= 0\\3x(x-3)(x+3)(2x^2+1) &= 0\\x &= -3, 0, 3

Intro Problem Revisit To find the lengths of the prism's sides, we need to factor 10x^3 - 25x^2 - 15x .

First, pull out the GCF among the three terms.

10x^3 - 25x^2 - 15x\\5x(2x^2 - 5x - 3)

Factor what is inside the parenthesis like a quadratic equation. ac = -6 and the factors of -6 that add up to -5 are -6 and 1.

5x(2x^2 - 5x - 3)= 5x(2x + 1)(x - 3)

Therefore, the lengths of the rectangular prism's sides are 5x , 2x + 1 , and x - 3 .

Guided Practice

Factor the following polynomials.

1. 3x^4+14x^2+8

2. 36x^4-25

3. Find all the real-number solutions of 8x^5+26x^3-24x = 0 .

Answers

1. ac = 24 and the factors of 24 that add up to 14 are 12 and 2.

& 3x^4+14x^2+8\\& 3x^4+12x^2+2x^2+8\\& 3x^2(x^2+4)+2(x^4+4)\\& (x^2+4)(3x^2+2)

2. Factor this polynomial like a difference of squares.

& 36x^4-25\\& (6x^2-5)(6x^2+5)

6 and 5 are not square numbers, so this cannot be factored further.

3. Pull out a 2x from each term.

8x^5+26x^3-24x &= 0\\2x(4x^4+13x-12) &= 0\\2x(4x^4+16x^2-3x^2-12) &= 0\\2x[4x^2(x^2+4)-3(x^2+4)] &= 0\\2x(x^2+4)(4x^2-3) &= 0

Set each factor equal to zero.

& \qquad \qquad \qquad \qquad \qquad \qquad \ \ 4x^2-3 = 0\\& 2x = 0 \quad x^2+4 = 0\\& \qquad \quad \ \ \ \ \quad \qquad \qquad \ \ and \qquad \quad x^2 = \frac{3}{4}\\& \ x = 0 \qquad \quad x^2 = -4\\& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ x = \pm \frac{\sqrt{3}}{2}

Notice the second factor will give imaginary solutions.

Vocabulary

Quadratic form
When a polynomial looks a trinomial or binomial and can be factored like a quadratic equation.

Practice

Factor the following quadratics completely.

  1. x^4-6x^2+8
  2. x^4-4x^2-45
  3. x^4-9x^2+45
  4. 4x^4-11x^2-3
  5. 6x^4+19x^2+8
  6. x^4-81
  7. 16x^4-1
  8. 6x^5+26x^3-20x
  9. 4x^6-36x^2
  10. 625-81x^4

Find all the real-number solutions to the polynomials below.

  1. 2x^4-5x^2-12 = 0
  2. x^4-16=0
  3. 16x^4-49 = 0
  4. 12x^6+69x^4+45x^2 = 0
  5. 3x^4+17x^2 -6=0

Image Attributions

Description

Difficulty Level:

At Grade

Grades:

Date Created:

Mar 12, 2013

Last Modified:

Sep 19, 2014

We need you!

At the moment, we do not have exercises for Factoring Polynomials in Quadratic Form.

Files can only be attached to the latest version of Modality

Reviews

Please wait...
Please wait...
Image Detail
Sizes: Medium | Original
 
MAT.ALY.212.L.2
ShareThis Copy and Paste

Original text