How do you factor #6x^3+10x^2-32#?

1 Answer
bp
Sep 30, 2015

#2 (x-4/3)(3x^2 +8x +12)#

Explanation:

After taking out 2 as a common factor it becomes #2(3x^3 +5x^2 -16)#

In this 3rd degree polynomial, the product of its roots (zeros) would be #-16/3#. Try #x=4/3#, it is found that it is on of the roots. Hence #(x-4/3)# would be a factor. Now carry out long or synthetic division of the polynomial by #(x-4/3)# to have #3x^2 + 8x+12# as a quotient.

The required factorisation would be thus #2 (x-4/3)(3x^2 +8x +12)#

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