Chapter 2, Problem 93CR

To determine

To calculate: zeros of the polynomial function and also write polynomials as a product of linear factors.

Answer to Problem 93CR

The zeros of the polynomial function will be at $x=2$ and $x=-\frac{3}{2}$ .

And product of linear factors is $\left(x+2\right)\left(2x+3\right)\left(x^2-2x+2\right)$ .

Explanation of Solution

Given information:

  $y=2x^4-5x^3+10x-12$

Formula used: To write the polynomial as a product of linear factors we first find the zeros of the polynomial and then combine the linear expressions that has those zeros.

From the linear equation $12$ is the constant term , therefore write the factors of $12$ is (p) and the variable with greatest power of x is (q) and then put $\frac{p}{q}$ value in the linear equation to get the zeros of the polynomial.

Plot the graph to show the polynomials.

  

Calculation: Factors of $12$ is $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12$ which is p

The possible values of q is $\pm 1,\pm 2$

Therefore $\frac{p}{q}=\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12,\pm \frac{1}{2},\pm \frac{3}{2}$

Now put these values in the polynomial to get the zeros,

  $f\left(1\right)=2\times 1^4-5\times 1^3+10\times 1-12=2-5+10-12=-5$

  $f\left(-1\right)=2\times -1^4-5\times -1^3+10\times -1-12=2+5-10-12=-15$

  $f\left(2\right)=2\times 2^4-5\times 2^3+10\times 2-12=32-40+20-12=0$

  $f\left(-2\right)=2\times -2^4-5\times -2^3+10\times -2-12=32+40-20-12=40$

  $f\left(3\right)=2\times 3^4-5\times 3^3+10\times 3-12=162-135+30-12=45$

  $f\left(-3\right)=2\times -3^4-5\times -3^3+10\times -3-12=162+135-30-12=285$

  $f\left(4\right)=2\times 4^4-5\times 4^3+10\times 4-12=512-320+40-12=220$

  $f\left(-4\right)=2\times -4^4-5\times -4^3+10\times -4-12=512+320-40-12=780$

  $f\left(6\right)=2\times 6^4-5\times 6^3+10\times 6-12=2592-1130+60-12=1510$

  $f\left(-6\right)=2\times -6^4-5\times -6^3+10\times -6-12=2592+1130-60-12=3650$

  $f\left(12\right)=2\times {12}^4-5\times {12}^3+10\times 12-12=41472-8640+120-12=32940$

  $f\left(-12\right)=2\times -{12}^4-5\times -{12}^3+10\times -12-12=41472+8640-120-12=49680$

  $f\left(\frac{1}{2}\right)=2\times {\left(\frac{1}{2}\right)}^4-5\times {\left(\frac{1}{2}\right)}^3+10\times \frac{1}{2}-12=\frac{1}{8}-\frac{5}{8}-7=-\frac{60}{8}$

  $f\left(-\frac{1}{2}\right)=2\times {\left(-\frac{1}{2}\right)}^4-5\times {\left(-\frac{1}{2}\right)}^3+10\times -\frac{1}{2}-12=\frac{1}{8}+\frac{5}{8}-17=-\frac{130}{8}$

  $f\left(\frac{3}{2}\right)=2\times {\left(\frac{3}{2}\right)}^4-5\times {\left(\frac{3}{2}\right)}^3+10\times \frac{3}{2}-12=\frac{81}{8}-\frac{135}{8}+3=-\frac{30}{8}$

  $f\left(-\frac{3}{2}\right)=2\times {\left(-\frac{3}{2}\right)}^4-5\times {\left(-\frac{3}{2}\right)}^3+10\times -\frac{3}{2}-12=\frac{81}{8}+\frac{135}{8}-27=0$

Hence we can observe that the zeros of the polynomial function will be at $x=2$ and $x=-\frac{3}{2}$ is shown in the graph also the function will be zero at that point.

If  $a$  is a root of the polynomial  $p\left(x\right)$ ,then the remainder from the division of  $p\left(x\right)$ by $x-a$ should equal 0

Since zeros will be at $x=2$ and $x=-\frac{3}{2}$ ,hence $x-2$ and $x+\frac{3}{2}$ is the factor of the polynomial

Therefore the product of linear factors are,

  $2x^4-5x^3+10x-12=\left(x-2\right)\left(2x^3-x^2-2x+6\right)$

  $\begin{array}{l}\Rightarrow \left(x+2\right)\left(x+\frac{3}{2}\right)\left(2\left(x^2-2x+2\right)\right)\\ \Rightarrow \left(x+2\right)\left(2x+3\right)\left(x^2-2x+2\right)\end{array}$