How do you factor completely: #x^2 + 5x + 6#?

2 Answers
Mar 13, 2018

Answer:

#(x + 3)(x + 2 )#

Explanation:

You need to first find 2 number which multiply to make 6 and add to make 5. It is easiest to list the factor pairs of 6:
#6 = 1 xx 6#
#6 = 2 xx 3#

#1 + 6 = 7#
#2 + 3 = 5#

Therefore we use the numbers 2 and 3. it doesnt matter which way round they go in the brackets.

Mar 13, 2018

Answer:

#(x+2)(x+3)#, if #y=0#, #x=-2# or #-3#

Explanation:

First, we need to find factors of 6. The factors of 6 are 1,2,3, and 6.
Now see the function. #x^2+5x+6#. Because there is no substract operation,
#(x+a)(x+b) = x^2+5x+6#
#bx+ax=5x,# and #ab=6#
#x(b+a)=5x#, so #b+a=5#
#a+b=5# and #ab=6#.
Look at the factors. a must be 2 or 3, let's say a is 2, and b must be 3.
So: #(x+2)(x+3)=0#, (to find the factor)
#x+2=0, x=-2 # ...(1)
#x+3=0, x=-3 # ... (2)