What is the standard form of $y = {(x - 5)}^{2} (6 - x)$ $y= (x - 5)^2(6-x)$ ?

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Answer 1 · 2015-12-17T03:45:33

$y = - x^{3} + 16 x^{2} - 50 x + 150$ $y=-x^3+16x^2-50x+150$

First expand ${(x - 5)}^{2}$ $(x-5)^2$ .

${(x - 5)}^{2} = (x - 5) (x - 5) = x^{2} - 10 x + 25$ $(x-5)^2=(x-5)(x-5)=x^2-10x+25$

Now, you know that $y = (x^{2} - 10 x + 25) (6 - x)$ $y=(x^2-10x+25)(6-x)$

Distribute again.

$y = 6 x^{2} - 60 x + 150 - x^{3} + 10 x^{2} - 25 x$ $y=6x^2-60x+150-x^3+10x^2-25x$

Combine like terms and write the terms in order of descending degree (biggest exponent first).

$y = - x^{3} + 16 x^{2} - 50 x + 150$ $y=-x^3+16x^2-50x+150$

What is the standard form of y=(x−5)2(6−x)y= (x - 5)^2(6-x)?