How do you express #f(x) = x^3+4x^2-5x-14# as a product of linear factors?

1 Answer
George C.
Aug 16, 2017

#f(x) = x^3+4x^2-5x-14#

#color(white)(f(x)) = (x-2)(x+3-sqrt(2))(x+3+sqrt(2))#

Explanation:

Given:

#f(x) = x^3+4x^2-5x-14#

Note that linear factors correspond to zeros of a polynomial. Explicitly, #(x-a)# is a factor if and only if #x=a# is a zero.

So if we find a zero then we find a factor.

By the rational root theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-14# and #q# a divosor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1, +-2, +-7, +-14#

We find:

#f(2) = (color(blue)(2))^3+4(color(blue)(2))^2-5(color(blue)(2))-14#

#color(white)(f(2)) = 8+16-10-14#

#color(white)(f(2)) = 0#

So #x=2# is a zero and #(x-2)# a factor:

#x^3+4x^2-5x-14 = (x-2)(x^2+6x+7)#

We can factor the remaining quadratic by completing the square and using the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with #a=(x+3)# and #b=sqrt(2)# as follows:

#x^2+6x+7 = x^2+6x+9-2#

#color(white)(x^2+6x+7) = (x+3)^2-(sqrt(2))^2#

#color(white)(x^2+6x+7) = ((x+3)-sqrt(2))((x+3)+sqrt(2))#

#color(white)(x^2+6x+7) = (x+3-sqrt(2))(x+3+sqrt(2))#

Putting it all together:

#f(x) = x^3+4x^2-5x-14#

#color(white)(f(x)) = (x-2)(x+3-sqrt(2))(x+3+sqrt(2))#

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