How do you factor #x^ { 2} + 6 x - 27= 0#?

2 Answers
Somebody N.
Sep 22, 2017

#(x - 3) ( x + 9)#

Explanation:

To factor #x^2 + 6x - 27 #

Find to numbers whose product is -27 and whose sum is 6. For this it is -3 and 9.

# -3 xx 9 = - 27# and #-3 + 9 = 6#.

Rewrite the equation as:

#x^2 - 3x + 9x -27#

Factor the first 2 terms:

#x(x - 3)#

Factor the last 2 terms:

#9(x - 3)#

So now we have:

#x(x -3) + 9(x -3)#

Notice we can bracket off this and factor out the expressions in parenthesis:

#[x(x-3) + 9(x-3)]#

#(x-3)[ x + 9 ]#

These are our required factors:

#(x - 3) ( x + 9)#

George C.
Sep 22, 2017

#(x+9)(x-3) = 0#

Explanation:

Given:

#x^2+6x-27#

Here are a couple of methods:

Method 1 - Fishing for factors

Find a pair of factors of #27# with difference #6#.

The pair #9, 3# works in that #9*3 = 27# and #9-3=6#

So we find:

#x^2+6x-27 = (x+9)(x-3)#

Method 2 - Completing the square

#x^2+6x-27 = x^2+6x+9-36#

#color(white)(x^2+6x-27) = (x+3)^2-6^2#

#color(white)(x^2+6x-27) = ((x+3)-6)((x+3)+6)#

#color(white)(x^2+6x-27) = (x-3)(x+9)#

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