How do you find all zeros of $f (x) = x^{4} - x^{3} - 20 x^{2}$ $f(x)=x^4-x^3-20x^2$ ?

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How do you find all zeros of $f (x) = x^{4} - x^{3} - 20 x^{2}$ $f(x)=x^4-x^3-20x^2$ ?

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Answer 1 · 2017-12-12T14:24:28

$x = - 4, x = 0, x = 5$ $x=-4,x=0,x=5$

Explanation:

$take out a common factor of x^{2}$ $"take out a common factor of "x^2$

$f (x) = x^{2} (x^{2} - x - 20)$ $f(x)=x^2(x^2-x-20)$

$the factors of - 20 which sum to - 1 are - 5 and + 4$ $"the factors of - 20 which sum to - 1 are - 5 and + 4"$

$\Rightarrow f (x) = x^{2} (x - 5) (x + 4)$ $rArrf(x)=x^2(x-5)(x+4)$

$to find zeros set f (x) = 0$ $"to find zeros set "f(x)=0$

$\Rightarrow x^{2} (x - 5) (x + 4) = 0$ $rArrx^2(x-5)(x+4)=0$

$equate each factor to zero and solve for x$ $"equate each factor to zero and solve for x"$

$x^{2} = 0 \Rightarrow x = 0 with multiplicity 2$ $x^2=0rArrx=0" with multiplicity 2"$

$x - 5 = 0 \Rightarrow x = 5$ $x-5=0rArrx=5$

$x + 4 = 0 \Rightarrow x = - 4$ $x+4=0rArrx=-4$

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How do you find all zeros of $f (x) = x^{4} - x^{3} - 20 x^{2}$ $f(x)=x^4-x^3-20x^2$ ?

1 Answer

Explanation:

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