How do you factor $- p^{2} + 4 p - 4 = 0$ $-p^{2}+4p-4=0$ ?

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How do you factor $- p^{2} + 4 p - 4 = 0$ $-p^{2}+4p-4=0$ ?

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Answer 1 · 2018-03-23T03:07:21

$- {(p - 2)}^{2} = 0$ $-(p-2)^2=0$
This involves factoring out the negative from the trinomial.

Explanation:

You want the leading coefficient on the $p^{2}$ $p^2$ to be positive. This means that you should factor out the negative, which will switch the signs.

$\Rightarrow$ $=>$ $- (p^{2} - 4 p + 4) = 0$ $-(p^2-4p+4)=0$

Now that the negative is factored out, you should factor the rest like normal. As $p^{2}$ $p^2$ has no coefficient, the $p$ $p$ in the factorization must have no coefficient. Seeing the $4$ $4$ 's, 2 should pop into your head, because both $2 + 2$ $2+2$ and $2 \cdot 2$ $2*2$ equal $4$ $4$ . The constant is positive, which implies a $- 2$ $-2$ , because $- 2^{2} = 4$ $-2^2=4$ .

These conclusions lead to:

$\Rightarrow$ $=>$ ${(p - 2)}^{2} = 0$ $(p-2)^2=0$

Now add the negative in front, for:

$\Rightarrow$ $=>$ $- {(p - 2)}^{2} = 0$ $-(p-2)^2=0$

You can test the answer by using FOIL:

$\Rightarrow$ $=>$ $- ((p - 2) (p - 2)) = 0$ $-((p-2)(p-2))=0$
$\Rightarrow$ $=>$ $- (p^{2} - 4 p + 4) = 0$ $-(p^2-4p+4)=0$

Distribute the negative:

$\Rightarrow$ $=>$ $- p^{2} + 4 p - 4 = 0$ $-p^2+4p-4=0$

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How do you factor $- p^{2} + 4 p - 4 = 0$ $-p^{2}+4p-4=0$ ?

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Explanation:

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How do you factor -p^{2}+4p-4=0−p2+4p−4=0-p^{2}+4p-4=0?

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How do you factor $- p^{2} + 4 p - 4 = 0$ $-p^{2}+4p-4=0$ ?