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Methods for Solving Quadratic Functions

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Factoring Polynomials in Quadratic Form

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 the Leading Coefficient Doesn't Equal 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^4-18x^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

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