Question #3ed52

1 Answer
Jan 9, 2017

Answer:

#x^2-4=(x+2)(x-2)#
#2x^2-x-10=(2x-5)(x+2)#
#5x+10=5(x+2)#

Explanation:

Starting with #5x+10#, we always start factoring by checking if there is a common factor, that is, a number or variable that divides into every term evenly.
Since both 5x and 10 can be divided by 5, we write the 5 outside brackets, and then write inside the brackets what we would get if we divided each term by 5. To check your answer, multiply using the distributive property. #color(white)(aaa)5x+10=5(x+2)#

For #2x^2-x-10#, the standard method used in this area is the method of decomposition.
#color(white)(aaaa)# First multiply the coefficients of the first and last terms: #color(white)(aaaaaaaa)2(-10)=-20#
#color(white)(aaaa)#Then look for factors of #-20# that add to the coefficient of the#color(white)(aaaaa)# middle term, #-1#.
#color(white)(aaaa)#The required factors are #4# and #-5# which multiply to #-20##color(white)(aaaaa)# and add to #-1#: #color(white)(aaaaaa)4-5=-1#

We now rewrite the question, decomposing the middle term:
#color(white)(aaaa)#Then, #color(white)(aaa)2x^2-x-10=2x^2+4x-5x-20#

Grouping two terms at a time, take out a common factor:
#color(white)(aaaaaaaaaaaaaaaaaaaaaa)=2x(x+4)-5(x+4)#

Since #(x+4)# is a common factor for the two terms we now have, take out #(x+4)# as a common factor:
#color(white)(aaaaaaaaaaaaaaaaaaaaaa)=(2x-5)(x+4)#

The first question, #x^2-4#, is a type of factoring called "difference of squares." Since each term is a perfect square, the question can be rewritten as #(x)^2-(2)^2#.
To factor this type of question, write two brackets and in the front of each bracket put the x, in the back of each bracket put the 2, and make the signs in the middle of each bracket different.
So, #color(white)(aaaa)x^2-4=(x+2)(x-2)#

The correctness of the factoring of each of these questions can be checked by multiplying the answer. The original question should result.