How do you find $p (x - 4)$ $p(x-4)$ given $p (a) = a^{3} - 5$ $p(a)=a^3-5$ ?

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Answer 1 · 2018-03-15T16:13:43

$p (x - 4) = x^{3} - 12 x^{2} + 48 x - 69$ $p(x - 4) = x^3 - 12x^2 + 48x - 69$

Replace $a$ $a$ with $x - 4$ $x - 4$ :

$p (a) = a^{3} - 5$ $p(a) = a^3 - 5$

$p (x - 4) = {(x - 4)}^{3} - 5$ $p(x - 4) = (x - 4)^3 - 5$

$p (x - 4) = x^{3} + 3 x^{2} (- 4) + 3 x {(- 4)}^{2} - 4^{3} - 5$ $p(x - 4) = x^3 + 3x^2(-4) + 3x(-4)^2 - 4^3 - 5$

$p (x - 4) = x^{3} - 12 x^{2} + 48 x - 69$ $p(x - 4) = x^3 - 12x^2 + 48x - 69$

How do you find p(x−4)p(x-4) given p(a)=a3−5p(a)=a^3-5?