Find the value of the polynomial $5 x-4 x^{2}+3$ at <br/> <br/>(i) $x=0$ <br/> <br/>(ii) $x=-1$ <br/> <br/>(iii) $x=2$
Solution:
(i) $p(x)=5 x-4 x^{2}+3$
$p(0)=5(0)-4(0)^{2}+3$
$=3$
(ii) $p(x)=5 x-4 x^{2}+3$
$p(-1)=5(-1)-4(-1)^{2}+3$
$=-5-4(1)+3=-6$
(iii) $p(x)=5 x-4 x^{2}+3$
$p(2)=5(2)-4(2)^{2}+3$
$=10-16+3=-3$
(i) $p(x)=5 x-4 x^{2}+3$
$p(0)=5(0)-4(0)^{2}+3$
$=3$
(ii) $p(x)=5 x-4 x^{2}+3$
$p(-1)=5(-1)-4(-1)^{2}+3$
$=-5-4(1)+3=-6$
(iii) $p(x)=5 x-4 x^{2}+3$
$p(2)=5(2)-4(2)^{2}+3$
$=10-16+3=-3$
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