How do you factor $4 p^{2} - 16 p - 84$ $4p ^ { 2} - 16p - 84$ ?

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How do you factor $4 p^{2} - 16 p - 84$ $4p ^ { 2} - 16p - 84$ ?

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Answer 1 · 2017-07-18T21:40:15

$4 p^{2} - 16 p - 84$ $4p^2-16p-84$

= $4 p^{2} - 16 p + 16 - 100$ $4p^2-16p+16-100$

= ${(2 p - 4)}^{2} - 10^{2}$ $(2p-4)^2-10^2$

= $(2 p - 4 + 10) \cdot (2 p - 4 - 10)$ $(2p-4+10)*(2p-4-10)$

= $(2 p + 6) \cdot (2 p - 14)$ $(2p+6)*(2p-14)$

Explanation:

I used difference of squares identity.

Answer 2 · 2017-07-18T23:08:13

$4 p^{2} - 16 p - 84 = 4 (p - 7) (p + 3)$ $4p^2-16p-84 = 4(p-7)(p+3)$

Explanation:

Given:

$4 p^{2} - 16 p - 84$ $4p^2-16p-84$

Note that all of the coefficients are divisible by $4$ $4$ , so we can separate that out as a constant factor.

The remaining quadratic is monic (leading coefficient $1$ $1$ ), so we then just need to find two factors of $21$ $21$ which differ by $4$ $4$ . The pair $7, 3$ $7, 3$ works, so we find:

$4 p^{2} - 16 p - 84 = 4 (p^{2} - 4 p - 21)$ $4p^2-16p-84 = 4(p^2-4p-21)$

$4 p^{2} - 16 p - 84 = 4 (p - 7) (p + 3)$ $color(white)(4p^2-16p-84) = 4(p-7)(p+3)$

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How do you factor $4 p^{2} - 16 p - 84$ $4p ^ { 2} - 16p - 84$ ?

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