How do you factor $p^{2} - 14 p + 49$ $p^2-14p+49$ using the perfect squares formula?

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How do you factor $p^{2} - 14 p + 49$ $p^2-14p+49$ using the perfect squares formula?

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Answer 1 · 2017-04-15T19:41:20

$(p - 7) (p - 7)$ $(p-7)(p-7)$

Explanation:

49 is $7^{2}$ $7^2$ so we need to make -14p and +49; therefore both signs in the brackets must minus.

$(p - 7) (p - 7)$ $(p-7)(p-7)$

Tip; The perfect square formula can be a bit confusing. Just use your intuition if you know the answer. That is what I do and I do not get penalised for it. Just use the formula if necessary.

Answer 2 · 2017-04-15T19:44:40

See below.

Explanation:

The perfect squares formula is in the form: ${(x - a)}^{2}$ $(x-a)^2$ . Expanding,

${(x - a)}^{2} = x^{2} - 2 a x + a^{2}$ $(x-a)^2=x^2-2ax+a^2$

Set this equal to $x^{2} - 14 x + 49$ $x^2-14x+49$ .

So:

$a^{2} = 49$ $a^2=49$ , or $a = \pm 7$ $a=\pm7$

But,

$- 2 a x = - 14 x$ $-2ax=-14x$ , so $a = 7$ $a=7$ .

Thus, $p^{2} - 14 p + 49 = {(p - 7)}^{2}$ $p^2-14p+49=(p-7)^2$ .

Or, factor by splitting.

$p^{2} - 7 p - 7 p + 49$ $p^2-7p-7p+49$

$p (p - 7) - 7 (p - 7)$ $p(p-7)-7(p-7)$

$(p - 7) (p - 7)$ $(p-7)(p-7)$

$= {(p - 7)}^{2}$ $=(p-7)^2$

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How do you factor $p^{2} - 14 p + 49$ $p^2-14p+49$ using the perfect squares formula?

2 Answers

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Explanation:

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How do you factor p^2-14p+49p2−14p+49p^2-14p+49 using the perfect squares formula?

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How do you factor $p^{2} - 14 p + 49$ $p^2-14p+49$ using the perfect squares formula?