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How do you find the zeroes of #p(x)= x^4-4x^3-x^2+16x-12#?

Answer 1

Cora Alexander

By observation we can note that if #p(x)=x^4-4x^3-x^2+16x-12# then #p(1) = 0#

Therefore #(x-1)# is a factor of #p(x)#

By synthetic division we obtain #p(x) = (x-1)(x^3-3x^2-4x+12)#

Focusing on #(x^3-3x^2-4x+12)# we notice the grouping #x^2(x-3) -4(x-3)#

#=(x^2-4)(x-3)#

and using the difference of squares #=(x+2)(x-2)(x-3)#

Therefore a complete factoring of #p(x)# #=(x-1)(x+2)(x-2)(x-3)#

and the zeroes of #p(x)# are #{1,-2, 2, 3}#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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