Question

Using factor theorem, factorize each of the following polynomials: $2y^3-5y^2-19y+42$ .

Answer 1

Hint: To factorize $2y^3-5y^2-19y+42$ using factor theorem, we have to equate the given polynomial to f(y). Then, we have a trial and error method by substituting some values of y so that the above function will become 0. If $$f(-c)=0$$ , then $$(y+c)$$ is a factor of the polynomial f(y). If $$f\left({\displaystyle \frac{d}{c}}\right)=0$$ then $$(cy-d)$$ is a factor of the polynomial f(y). If $$f(-{\displaystyle \frac{d}{c}})=0$$ , then $$(cy+d)$$ is a factor of the polynomial f(y). . If $$f\left(c\right)=0$$ , then $$(y-c)$$ is a factor of the polynomial f(y). Then, we will divide the given polynomial by this factor. The quotient will be another polynomial. If we get the quotient as a quadratic polynomial, we have to further factorize this polynomial using splitting the middle terms or using the formula $y={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$ .

Complete step by step solution:
We have to factorize $2y^3-5y^2-19y+42$ using the factor theorem. We have to first equate the given polynomial to f(y).
$\Rightarrow f\left(y\right)=2y^3-5y^2-19y+42$
Now, we have to substitute some values of y so that the above function will become 0. This is the trial and error method.
Let us consider $y=0$ .
$\Rightarrow f\left(0\right)=2\times 0-5\times 0-19\times 0+42=42\ne 0$
Hence, this value of y cannot be taken.
Now, we have to consider $y=1$ .
$\begin{array}{rl}& \Rightarrow f\left(1\right)=2\times 1-5\times 1-19\times 1+42\\ & \Rightarrow f\left(1\right)=2-5-19+42=20\ne 0\end{array}$
Hence, this value of y cannot be taken.
Now, let us consider $y=2$
$\begin{array}{rl}& \Rightarrow f\left(2\right)=2{\left(2\right)}^3-5{\left(2\right)}^2-19\times 2+42\\ & \Rightarrow f\left(2\right)=16-20-38+42=0\end{array}$
We can see that $(y-2)$ is a factor of $2y^3-5y^2-19y+42$ .
Now, we have to divide $2y^3-5y^2-19y+42$ by $(y-2)$ .
$$y-2\stackrel{2y^2-y-21}{\overline{)\begin{array}{rl}& 2y^3-5y^2-19y+42\\ & 2y^3-4y^2\\ & (-)\begin{array}{cc}& (+)\end{array}\\ & \mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\\ & \begin{array}{ccc}& & -y^2-19y\end{array}\\ & \begin{array}{ccc}& & -y^2+2y\end{array}\\ & \begin{array}{ccc}& & (+)\begin{array}{cc}& (-)\end{array}\end{array}\\ & \mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\\ & \begin{array}{ccc}& & \begin{array}{ccc}& & -21y+42\end{array}\end{array}\\ & \begin{array}{ccc}& & \begin{array}{ccc}& & -21y+42\end{array}\end{array}\\ & \begin{array}{ccc}& & \begin{array}{ccc}& & (+)\begin{array}{cc}& (-)\end{array}\end{array}\end{array}\\ & \mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\\ & \begin{array}{ccc}& & \begin{array}{ccc}& & \begin{array}{ccc}& & 0\end{array}\end{array}\end{array}\end{array}}}$$
Hence, we can write f(y) as
$f\left(y\right)=(y-2)(2y^2-y-21)$
Now, we can factorize $2y^2-y-21$ by splitting the middle term. We can write –y as $$6y-7y$$ as the sum is -1 and product is -21.
$\Rightarrow f\left(y\right)=(y-2)(2y^2+6y-7y-21)$
Let us take the common factors outside.
$\Rightarrow f\left(y\right)=(y-2)[2y(y+3)-7(y+3)]$
We can again take the common factors outside.
$\Rightarrow f\left(y\right)=(y-2)(2y-7)(y+3)$

Note: Students must know the long division method thoroughly to factorize the given polynomial. If $$f(-c)=0$$ , then $$(y+c)$$ is a factor of the polynomial f(y). If $$f\left({\displaystyle \frac{d}{c}}\right)=0$$ then $$(cy-d)$$ is a factor of the polynomial f(y). If $$f(-{\displaystyle \frac{d}{c}})=0$$ , then $$(cy+d)$$ is a factor of the polynomial f(y). . If $$f\left(c\right)=0$$ , then $$(y-c)$$ is a factor of the polynomial f(y). We can also use synthetic division instead of long division.