How do you factor the trinomial #8z^2+20z-48#?

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Apr 22, 2018

Answer:

#4(z+4)(2z-3)#

Explanation:

you can divide all the terms by a common factor #4# to get

#2z^2 + 5z - 12#

this expression can be factorised by grouping:

the first and last coefficients are #2# and #-12#.
the product of this is #-24#.

then, you can find #2# numbers that add to give the second coefficient, and multiply to give the product of the first and last coefficient.

#8 + -3 = 8 - 3 = 5#
#8 * -3 = -24#

the two numbers that add to give the second coefficient are #8# and #-3#.
therefore the second term can be expressed as the sum of two terms, #8z# and #-3z#.

this gives the expression #2z^2 - 3z + 8z - 12#.

then you can factorise each adjacent pair:

#2z^2 - 3z = z(2z-3)#
#8z - 12 = 4(2z-3)#

therefore #2z^2 - 3z + 8z - 12# can be written as #z(2z-3) + 4(2z-3)#.
this sum is the same as #(z+4)(2z-3)#.

then you have the expression #2z^2+5z-12# grouped as #(z+4)(2z-3)#.

the expression given in the question is #8z^2+20z-48#.
that is #4# times #2z^2+5z-12#, or the grouped expression #(z+4)(2z-3)#.
therefore it can be written as #4(z+4)(2z-3)#.

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