2.1.1: Methods for Solving Quadratic Functions
- Page ID
- 14122
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Factoring Polynomials in Quadratic Form
The volume of a rectangular prism is 10x3−25x2−15x. What are the lengths of the prism's sides?
Factoring Polynomials in Quadratic Form
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 ax4+bx2+c. Another possibility is something similar to the difference of squares, a4−b4. This can be factored to (a2−b2)(a2+b2) or (a−b)(a+b)(a2+b2). Always keep in mind that the greatest common factors should be factored out first.
Let's factor the following polynomials.
- 2x4−x2−15
This particular polynomial is factorable. 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.
2x4−x2−15
2x4−6x2+5x2−15
2x2(x2−3)+5(x2−3)
(x2−3)(2x2+5)
Both of the factors are not factorable, so we are done.
- 81x4−16
Treat this polynomial equation like a difference of squares.
81x4−16
(9x2−4)(9x2+4)
Now, we can factor 9x2−4 using the difference of squares a second time.
(3x−2)(3x+2)(9x2+4)
9x2+4 cannot be factored because it is a sum of squares. This will have imaginary solutions.
Now, let's find all the real-number solutions of 6x5−51x3−27x=0.
First, pull out the GCF among the three terms.
6x5−51x3−27x=0
3x(2x4−17x2−9)=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.
6x5−51x3−27x=0
3x(2x4−17x2−9)=0
3x(2x4−18x2+x2−9)=0
3x[2x2(x2−9)+1(x2−9)]=0
3x(x2−9)(2x2+1)=0
Factor x2−9 further and solve for x where possible. 2x2+1 is not factorable.
3x(x2−9)(2x2+1)=0
3x(x−3)(x+3)(2x2+1)=0
x=−3,0,3
Examples
Earlier, you were asked to find the lengths of the prism's sides.
Solution
To find the lengths of the prism's sides, we need to factor 10x3−25x2−15x.
First, pull out the GCF among the three terms.
10x3−25x2−15x
5x(2x2−5x−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(2x2−5x−3)=5x(2x+1)(x−3)
Therefore, the lengths of the rectangular prism's sides are 5x, 2x+1, and x−3.
Factor: 3x4+14x2+8.
Solution
ac=24 and the factors of 24 that add up to 14 are 12 and 2.
3x4+14x2+8
3x4+12x2+2x2+8
3x2(x2+4)+2(x4+4)
(x2+4)(3x2+2)
Factor: 36x4−25.
Solution
Factor this polynomial like a difference of squares.
36x4−25
(6x2−5)(6x2+5)
6 and 5 are not square numbers, so this cannot be factored further.
Find all the real-number solutions of 8x5+26x3−24x=0.
Solution
Pull out a 2x from each term.
8x5+26x3−24x=0
2x(4x4+13x−12)=0
2x(4x4+16x2−3x2−12)=0
2x[4x2(x2+4)−3(x2+4)]=0
2x(x2+4)(4x2−3)=0
Set each factor equal to zero.
4x2−3=0
2x=0
x2+4=0
and x2= \(\ 3 \over 4\)
x=0
x2=−4
x=± \(\ \frac{\sqrt{3}}{2}\)
Notice the second factor will give imaginary solutions.
Review
Factor the following quadratics completely.
- x4−6x2+8
- x4−4x2−45
- x4−18x2+45
- 4x4−11x2−3
- 6x4+19x2+8
- x4−81
- 16x4−1
- 6x5+26x3−20x
- 4x6−36x2
- 625−81x4
Find all the real-number solutions to the polynomials below.
- 2x4−5x2−12=0
- x4−16=0
- 16x4−49=0
- 12x6+69x4+45x2=0
- 3x4+17x2−6=0
Vocabulary
Term | Definition |
---|---|
Factor to Solve | "Factor to Solve" is a common method for solving quadratic equations accomplished by factoring a trinomial into two binomials and identifying the values of x that make each binomial equal to zero. |
factored form | The factored form of a quadratic function f(x) is f(x)=a(x−r1)(x−r2), where r1 and r2 are the roots of the function. |
Factoring | Factoring is the process of dividing a number or expression into a product of smaller numbers or expressions. |
Quadratic form | A polynomial in quadratic form looks like a trinomial or binomial and can be factored like a quadratic expression. |
quadratic function | A quadratic function is a function that can be written in the form f(x)=ax2+bx+c, where a, b, and c are real constants and a≠0. |
Roots | The roots of a function are the values of x that make y equal to zero. |
standard form | The standard form of a quadratic function is f(x)=ax2+bx+c. |
Vertex form | The vertex form of a quadratic function is y=a(x−h)2+k, where (h,k) is the vertex of the parabola. |
Zeroes of a Polynomial | The zeroes of a polynomial f(x) are the values of x that cause f(x) to be equal to zero. |