Chapter 2, Problem 94CR

To determine

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

Answer to Problem 94CR

The zeros of the polynomial function will be at $x=-1$ and $x=\frac{1}{3}$

And product of linear factors is $\left(x+1\right)\left(x-\frac{1}{3}\right)\left(3x^2-6x+12\right)$

Explanation of Solution

Given information: $g\left(x\right)=3x^4-4x^3+7x^2+10x-4$

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 4 is the constant term , therefore write the factors of 4 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 4is $\pm 1,\pm 2,\pm 4$ which is p

The possible values of q is $\pm 1,\pm 3$

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

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

  $\begin{array}{l}f\left(1\right)=3\times 1^4-4\times 1^3+7\times 1^2+10\times 1-4=3-4+7+10-4=12\\ f\left(-1\right)=3\times -1^4-4\times -1^3+7\times -1^2+10\times -1-4=3+4+7-10-4=0\\ f\left(2\right)=3\times 2^4-4\times 2^3+7\times 2^2+10\times 2-4=48-32+28+20-4=60\\ f\left(-2\right)=3\times -2^4-4\times -2^3+7\times -2^2+10\times -2-4=48+32+28-20-4=84\\ f\left(4\right)=3\times 4^4-4\times 4^3+7\times 4^2+10\times 4-4=768-256+112+40-4=660\\ f\left(-4\right)=3\times -4^4-4\times -4^3+7\times -4^2+10\times -4-4=768+256+112-40-4=1092\end{array}$

  $\begin{array}{l}f\left(\frac{1}{3}\right)=3\times {\frac{1}{3}}^4-4\times {\frac{1}{3}}^3+7\times {\frac{1}{3}}^2+10\times \frac{1}{3}-4=\frac{1}{27}-\frac{4}{27}+\frac{7}{9}+\frac{10}{3}-4=0\\ f\left(-\frac{1}{3}\right)=3\times -{\frac{1}{3}}^4-4\times -{\frac{1}{3}}^3+7\times -{\frac{1}{3}}^2+10\times -\frac{1}{3}-4=\frac{1}{27}+\frac{4}{27}-\frac{7}{9}-\frac{10}{3}-4=-\frac{214}{27}\\ f\left(\frac{2}{3}\right)=3\times {\frac{2}{3}}^4-4\times {\frac{2}{3}}^3+7\times {\frac{2}{3}}^2+10\times \frac{2}{3}-4=\frac{16}{27}-\frac{32}{27}+\frac{28}{9}+\frac{20}{3}-4=\frac{140}{27}\\ f\left(-\frac{2}{3}\right)=3\times -{\frac{2}{3}}^4-4\times -{\frac{2}{3}}^3+7\times -{\frac{2}{3}}^2+10\times -\frac{2}{3}-4=\frac{16}{27}+\frac{32}{27}+\frac{28}{9}-\frac{20}{3}-4=-\frac{156}{27}\\ f\left(\frac{4}{3}\right)=3\times {\frac{4}{3}}^4-4\times {\frac{4}{3}}^3+7\times {\frac{4}{3}}^2+10\times \frac{4}{3}-4=\frac{256}{27}-\frac{256}{27}+\frac{112}{9}+\frac{40}{3}-4=\frac{196}{9}\\ f\left(-\frac{4}{3}\right)=3\times -{\frac{4}{3}}^4-4\times -{\frac{4}{3}}^3+7\times -{\frac{4}{3}}^2+10\times -\frac{4}{3}-4=\frac{256}{27}+\frac{256}{27}+\frac{112}{9}-\frac{40}{3}-4=\frac{380}{27}\end{array}$.

Hence we can observe that the zeros of the polynomial function will be at $x=-1$ and $x=\frac{1}{3}$ 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 $x=-1$ and $x=\frac{1}{3}$ , hence $x+1$ and $x-\frac{1}{3}$ is the factor of the polynomial

Therefore the product of linear factors are ,

  $\begin{array}{l}3x^4-4x^3+7x^2+10x-4=\left(x+1\right)\left(3x^3-7x^2+14x-4\right)\\ \Rightarrow \left(x+1\right)\left(x-\frac{1}{3}\right)\left(3x^2-6x+12\right)\\ \Rightarrow \left(x+1\right)\left(\frac{3x-1}{3}\right)\left(3\left(x^2-2x+4\right)\right)\\ \Rightarrow \left(x+1\right)\left(3x-1\right)\left(x^2-2x+4\right)\end{array}$