What is the standard form of $y=-(x - 1) (x^2 +6)(x + 5)^2$ ?

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Answer 1 · 2018-02-05T06:45:08

$y = -x^5 - 9x^4 -21x^3 -29x^2 - 90x + 150$

Given: $y = -(x-1)(x^2+6)(x+5)^2$

Standard form of a polynomial requires distribution and putting terms in descending order:

Note: $(a +b)^2 = (a^2 + 2ab + b^2)$

$y = -(x-1)(x^2+6)(x+5)^2$
$= -(x-1)(x^2+6)(x^2+10x+25)$

$y = -(x-1)(x^4 +10x^3 +25x^2 + 6x^2 +60x + 150)$

Add like terms:
$y = -(x-1)(x^4 + 10x^3 + 31x^2+60x + 150)$

Distribute again:

$y = -(x^5+ 10x^4 + 31x^3+60x^2 + 150x -x^4-10x^3-31x^2-60x-150)$

Add/Subtract like terms:

$y = -(x^5 + 9x^4 + 21x^3 + 29x^2 + 90x - 150)$

Distribute the negative sign:

$y = -x^5 - 9x^4 -21x^3 -29x^2 - 90x + 150$

What is the standard form of y=-(x - 1) (x^2 +6)(x + 5)^2 y=-(x - 1) (x^2 +6)(x + 5)^2 ?