SOLUTION: Find the maximum number of real zeros for the equation f(x)=x^3+7x^2+8x-16

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Question 1169630: Find the maximum number of real zeros for the equation f(x)=x^3+7x^2+8x-16
Found 2 solutions by josgarithmetic, MathLover1:
Answer by josgarithmetic(39613) About Me  (Show Source):
You can put this solution on YOUR website!
Three at the most. (degree of the variable is 3).

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

f%28x%29=x%5E3%2B7x%5E2%2B8x-16.........factor completely, write 7x%5E2 as+-x%5E2%2B8x%5E2+and 8x as -8x%2B16x
f%28x%29=x%5E3-x%5E2%2B8x%5E2-8x%2B16x-16...group
f%28x%29=%28x%5E3-x%5E2%29%2B%288x%5E2-8x%29%2B%2816x-16%29 ...factor out common
f%28x%29=x%5E2%28x-1%29%2B8x%28x-1%29%2B16%28x-1%29
f%28x%29=%28x+-+1%29+%28x%5E2+%2B+8+x+%2B+16%29
f%28x%29=%28x+-+1%29+%28x%5E2+%2B+8+x+%2B+4%5E2%29+........use square of sum rule
+f%28x%29=%28x+-+1%29+%28x+%2B+4%29%5E2
real zeros are:
%28x+-+1%29+%28x+%2B+4%29%5E2=0
if %28x+-+1%29=0 => highlight%28x=1%29
if %28x+%2B+4%29%5E2=0=> highlight%28x=-4%29-> double root

+graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+10%2C+x%5E3%2B7x%5E2%2B8x-16%29+