How do you factor the trinomial #b^2-4b-32#?

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Answer 1 · 2017-06-18T20:25:06

See a solution process below:

We can use the quadratic formula to factor this expression as:

The quadratic formula states:

For #ax^2 + bx + c = 0#, the values of #x# which are the solutions to the equation are given by:

#x = (-b +- sqrt(b^2 - 4ac))/(2a)#

Substituting #1# for #a#; #-4# for #b# and #-32# for #c# gives:

#b = (-(-4) +- sqrt((-4)^2 - (4 * 1 * -32)))/(2 * 1)#

#b = (4 +- sqrt(16 - (-128)))/(2)#

#b = (4 +- sqrt(16 + 128))/(2)#

#b = (4 +- sqrt(144))/(2)#

#b = (4 +- 12)/(2)#

#b = (4 + 12)/(2)# or #b = (4 - 12)/(2)#

#b = 16/2 or #b = -8/2#

#b = 8# or #b = -4#

#b - 8 = 0# or #b + 4 = 0#

#b^2 − 4b − 32 => (b - 8)(b + 4)#