Using factor theorem, show that g(x) is a factor of p(x), when

Question:

Using factor theorem, show that g(x) is a factor of p(x), when

$p(x)=2 x^{4}+x^{3}-8 x^{2}-x+6, g(x)=2 x-3$

 

Solution:

Let:

$p(x)=2 x^{4}+x^{3}-8 x^{2}-x+6$

Here, 

$2 x-3=0 \Rightarrow x=\frac{3}{2}$

By the factor theorem, $(2 x-3)$ is a factor of the given polynomial if $p\left(\frac{3}{2}\right)=0$.

Thus, we have:

$p\left(\frac{3}{2}\right)=\left[2 \times\left(\frac{3}{2}\right)^{4}+\left(\frac{3}{2}\right)^{3}-8 \times\left(\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)+6\right]$

$=\left(\frac{81}{8}+\frac{27}{8}-18-\frac{3}{2}+6\right)$

$=0$

Hence, (2x -">- 3) is a factor of the given polynomial.

 

Leave a comment

Comments

Mohisha
April 18, 2023, 11:42 a.m.
P(x) =x^4-x^2-12,g(x)=x+2
Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now