How do you factor and solve #x^2+8x+15=0#?

2 Answers

Answer:

Find roots that when added come to the middle term coefficient and when multiplied come to the last term and you'll find that #x=-3# and #x=-5#

Explanation:

The key to factoring these kind of quadratic equations is to find 2 roots that when added up equal the coefficient of the middle term and when multiplied equal the last term.

If I'm having problems finding it, I often list out the factors of the last term and see what adds up to the middle term, like this:

#1+15=16#
#3+5=8#

(when things get complicated with negative signs and coefficients in front of the first term, this method can really help keep things organized).

Ok - we know the roots are 3 and 5:

#(x+3)(x+5)=0#

The reason we set the left side equal to 0 is because it's easy to multiply by 0 in one term and get 0 as the overall answer. So what we do now is find out what it'll take to make x+3 = 0 and what it'll take x+5 = 0.

Let's do the x+3 first:

#x+3=0#
#x=-3#

Now the x+5 term:

#x+5=0#
#x=-5#

And those are your answers.

May 19, 2016

Answer:

#(x+3)(x+5)=0#
#color(white)("XXX")rarr (x=-3) or (x=-5)#

Explanation:

Since #3xx5=15# and #3+5=8#

#x^2+8x+15# can be factored as #(x+3)(x+5)#

And since #x^2+8x+15=0#
#color(white)("XXX")rArr (x+3)(x+5)=0#

#color(white)("XXX")rArr# either #(x+3)=0color(white)("XX")rarrcolor(white)("XX")x=-3#
#color(white)("XXXXXx")#or #color(white)("XX")(x+5)=0color(white)("XX")rarrcolor(white)("XX")x=-5#