How do you evaluate P(-1/2) if $P (x) = 2 x^{4} + x^{3} + 12$ $P(x) = 2x^4 + x^3 + 12$ ?

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How do you evaluate P(-1/2) if $P (x) = 2 x^{4} + x^{3} + 12$ $P(x) = 2x^4 + x^3 + 12$ ?

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Answer 1 · 2018-05-14T20:00:57

$P (- \frac{1}{2}) = 12$ $P(-1/2)=12$

Explanation:

$to evaluate P (- \frac{1}{2})$ $"to evaluate "P(-1/2)$

$substitute x = - \frac{1}{2} into P (x)$ $"substitute "x=-1/2" into "P(x)$

$P (- \frac{1}{2}) = 2 {(- \frac{1}{2})}^{4} + {(- \frac{1}{2})}^{3} + 12$ $P(color(red)(-1/2))=2(color(red)(-1/2))^4+(color(red)(-1/2))^3+12$

$\times \times \times = (2 \times \frac{1}{16}) - \frac{1}{8} + 12$ $color(white)(xxxxxx)=(2xx1/16)-1/8+12$

$color(white)(xxxxxx)=cancel(1/8)cancel(-1/8)+12=12$

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How do you evaluate P(-1/2) if $P (x) = 2 x^{4} + x^{3} + 12$ $P(x) = 2x^4 + x^3 + 12$ ?

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How do you evaluate P(-1/2) if P(x) = 2x^4 + x^3 + 12P(x)=2x4+x3+12P(x) = 2x^4 + x^3 + 12?

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How do you evaluate P(-1/2) if $P (x) = 2 x^{4} + x^{3} + 12$ $P(x) = 2x^4 + x^3 + 12$ ?