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
Comments
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.