Question

Simplify the following expressions and evaluate them as directed:
(i) $x^2(3-2x+x^2)$ for x=1;x=-1;$x=\frac{2}{3}andx=-\frac{1}{2}.$

Answer 1

$f\left(x\right)=x^2(3-2x+x^2)$

$x=1$

$f\left(x\right)=1^2[3-2(1)+(1{)}^2]$

$\Rightarrow 1[3-2+1]$

$\Rightarrow 2$

$x=-1$

$f\left(x\right)=(-1{)}^2[3-2(-1)+(-1{)}^2]$

$\Rightarrow 1[3+2+1]$

$\Rightarrow 6$

$x=\frac{2}{3}$

$f\left(x\right)=\left(\frac{2}{3}{)}^2\right[3-2\left(\frac{2}{3}\right)+\left(\frac{2}{3}{)}^2\right]$

$\Rightarrow \frac{4}{9}[3-\frac{4}{3}+\frac{4}{9}]$

$\Rightarrow \frac{4}{9}\left[\frac{27-12+4}{9}\right]\Rightarrow \frac{4}{9}\times \frac{19}{9}\Rightarrow \frac{76}{81}$

$x=-1/2$

$f\left(x\right)=(-\frac{1}{2}{)}^2[3-2(-\frac{1}{2})+(-\frac{1}{2}{)}^2]$

$\Rightarrow \frac{1}{4}[3+1+\frac{1}{4}]\Rightarrow \frac{1}{4}\left[\frac{12+4+1}{4}\right]\Rightarrow \frac{17}{16}$