Question

Find the value of
$p(x) = {x}^{3} + 2 {x}^{2}- 7x + 3$
at
$x = \frac{1}{2}$
and
$x = - 3$
Plz guys give the answer step by step. Plz don't write sorry or don't know plz. If whosoever writes like that I will name them as idiots. ​

Answer 1

Answer:

how can i know so taffff ap khod karlo

Step-by-step explanation:

u can do it

Answer 2

When x=1/2

$p(\frac{1}{2})=(\frac{1}{2})^3+2(\frac{1}{2})^2-7(\frac{1}{2}+3 \\ \\ =\frac{1}{8}+2(\frac{1}{4})-\frac{7}{2}+3 \\ \\ =\frac{1}{8}+\frac{2}{4}-\frac{7}{2}+3 \\ \\ =\frac{1+4-28+24}{8} \\ \\=\frac{1}{8}$

When x=-3

$p(-3)=(-3)^3+2(-3)^2-7(-3)+3 \\ =-27+2(9)+21+3 \\ =-27+18+21+3 \\ =15$

Hope this helps you