Asked by Natasha
find the real zeros of f(x)=x^4+6x^3-11x^2-24x+28
Answers
Answered by
Reiny
try x=1
f(1) = 1+6-11-24+28 = 0
so x-1 is a factor
do synthetic division
(x^4+6x^3-11x^2-24x+28) = (x-1)(x^3 + 7x - 4 - 28)
x^3 + 7x^2 - 4x - 28
= x^2(x+7) - 4(x+7)
= (x+7)(x^ - 4)
so x^4+6x^3-11x^2-24x+28
= (x-1)(x-2)(x+2)(x+7)
x = 1, -2, 2, -7
f(1) = 1+6-11-24+28 = 0
so x-1 is a factor
do synthetic division
(x^4+6x^3-11x^2-24x+28) = (x-1)(x^3 + 7x - 4 - 28)
x^3 + 7x^2 - 4x - 28
= x^2(x+7) - 4(x+7)
= (x+7)(x^ - 4)
so x^4+6x^3-11x^2-24x+28
= (x-1)(x-2)(x+2)(x+7)
x = 1, -2, 2, -7
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