How do you evaluate the function $p (x) = \frac{1}{2} x^{3} + \frac{2}{3} x^{2} - \frac{1}{4} x + \frac{1}{3}$ $p(x)=1/2x^3 + 2/3x^2 - 1/4x + 1/3$ for $p (- 2)$ $p(-2)$ ?

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How do you evaluate the function $p (x) = \frac{1}{2} x^{3} + \frac{2}{3} x^{2} - \frac{1}{4} x + \frac{1}{3}$ $p(x)=1/2x^3 + 2/3x^2 - 1/4x + 1/3$ for $p (- 2)$ $p(-2)$ ?

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Answer 1 · 2018-03-24T21:21:56

$p (- 2) = \frac{1}{2} {(- 2)}^{3} + \frac{2}{3} {(- 2)}^{2} - \frac{1}{4} (- 2) + \frac{1}{3}$ $p(-2)=1/2(-2)^3+2/3(-2)^2-1/4(-2)+1/3$
$= \frac{1}{2} (- 8) + \frac{2}{3} (4) + \frac{2}{4} + \frac{1}{3}$ $=1/2(-8)+2/3(4)+2/4+1/3$
$= - \frac{8}{2} + \frac{8}{3} + \frac{2}{4} + \frac{1}{3}$ $=-8/2+8/3+2/4+1/3$
$= - 4 + (\frac{8}{3} + \frac{1}{3}) + \frac{1}{2}$ $=-4+(8/3+1/3)+1/2$
$= - 4 + \frac{9}{3} + \frac{1}{2}$ $=-4+9/3+1/2$
$= - 4 + 3 + \frac{1}{2}$ $=-4+3+1/2$
$= - 1 + \frac{1}{2}$ $=-1+1/2$
$= - \frac{2}{2} + \frac{1}{2}$ $=-2/2+1/2$
$= \frac{- 2 + 1}{2}$ $=(-2+1)/2$
$= - \frac{1}{2}$ $=-1/2$

Explanation:

Just put the $x = - 2$ $x=-2$ on the polynomial equation and you will simplify and you will get $p (- 2) = - \frac{1}{2}$ $p(-2)=-1/2$

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How do you evaluate the function $p (x) = \frac{1}{2} x^{3} + \frac{2}{3} x^{2} - \frac{1}{4} x + \frac{1}{3}$ $p(x)=1/2x^3 + 2/3x^2 - 1/4x + 1/3$ for $p (- 2)$ $p(-2)$ ?

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How do you evaluate the function $p (x) = \frac{1}{2} x^{3} + \frac{2}{3} x^{2} - \frac{1}{4} x + \frac{1}{3}$ $p(x)=1/2x^3 + 2/3x^2 - 1/4x + 1/3$ for $p (- 2)$ $p(-2)$ ?