SOLUTION: How can I determine the zeros roots and interceptors of this third-degree polynomial function f ( x ) = x ^ 3 + x ^ 2-4x

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Question 984208: How can I determine the zeros roots and interceptors of this third-degree polynomial function f ( x ) = x ^ 3 + x ^ 2-4x
Found 2 solutions by Boreal, MathLover1:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
x^3+x^2-4x=0 to find roots
factor out x
x(x^2+x-4)=0
x=0 is one root
x=(1/2)(-1+/- sqrt (1+16)) That works ;;;; sqrt 17=4.12
x=1.56, -2.56
y intercept =0,0
graph%28300%2C200%2C-5%2C5%2C-10%2C10%2Cx%5E3%2Bx62-4x%29

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
first make f+%28+x+%29=0
0=+x%5E3+%2B+x%5E2-4x...factor out x
0+=+x%28x%5E2%2Bx-4%29
0+=+highlight%28x%29highlight%28%28x%5E2%2Bx-4%29%29... this product will be equal to zero only if one or both of the factors is equal to zero
so, one root is highlight%28x%5B1%5D=0%29
the other two roots we will get solving x%5E2%2Bx-4=0-use quadratic formula
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-1+%2B-+sqrt%28+1%5E2-4%2A1%2A%28-4%29+%29%29%2F%282%2A1%29+
x+=+%28-1+%2B-+sqrt%28+1%2B16+%29%29%2F2+
x+=+%28-1+%2B-+sqrt%28+17+%29%29%2F2+
exact solutions are:
highlight%28x%5B2%5D+=+%28-1+%2B+sqrt%28+17+%29%29%2F2%29+
and
highlight%28x%5B3%5D+=+%28-1+-+sqrt%28+17+%29%29%2F2%29+