How do you factor $a^{2} + 8 a - 48$ $a^2+8a-48$ ?

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How do you factor $a^{2} + 8 a - 48$ $a^2+8a-48$ ?

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Answer 1 · 2017-02-11T00:14:15

$a^{2} + 8 a - 48 = (a - 4) (a + 12)$ $a^2+8a-48 = (a-4)(a+12)$

Explanation:

The difference of squares identity can be written:

$A^{2} - B^{2} = (A - B) (A + B)$ $A^2-B^2 = (A-B)(A+B)$

Complete the square and use this with $A = (a + 4)$ $A=(a+4)$ and $B = 8$ $B=8$ as follows:

$a^{2} + 8 a - 48 = a^{2} + 8 x + 16 - 64$ $a^2+8a-48 = a^2+8x+16-64$

$a^{2} + 8 a - 48 = {(a + 4)}^{2} - 8^{2}$ $color(white)(a^2+8a-48) = (a+4)^2-8^2$

$a^{2} + 8 a - 48 = ((a + 4) - 8) ((a + 4) + 8)$ $color(white)(a^2+8a-48) = ((a+4)-8)((a+4)+8)$

$a^{2} + 8 a - 48 = (a - 4) (a + 12)$ $color(white)(a^2+8a-48) = (a-4)(a+12)$

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How do you factor $a^{2} + 8 a - 48$ $a^2+8a-48$ ?

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How do you factor $a^{2} + 8 a - 48$ $a^2+8a-48$ ?