How do you find all zeros for $(x + 1) (x - 4) (x^{2} - 7)$ $(x + 1)(x - 4)(x^2 - 7)$ ?

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How do you find all zeros for $(x + 1) (x - 4) (x^{2} - 7)$ $(x + 1)(x - 4)(x^2 - 7)$ ?

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Answer 1 · 2018-04-17T14:15:48

$x = - 1, x = 4, x = \pm \sqrt{7}$ $x=-1,x=4,x=+-sqrt7$

Explanation:

$to find the zeros equate the product of factors to zero$ $"to find the zeros equate the product of factors to zero"$

$\Rightarrow (x + 1) (x - 4) (x^{2} - 7) = 0$ $rArr(x+1)(x-4)(x^2-7)=0$

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

$x + 1 = 0 \Rightarrow x = - 1$ $x+1=0rArrx=-1$

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

$x^{2} - 7 = 0 \Rightarrow x^{2} = 7 \Rightarrow x = \pm \sqrt{7}$ $x^2-7=0rArrx^2=7rArrx=+-sqrt7$

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How do you find all zeros for $(x + 1) (x - 4) (x^{2} - 7)$ $(x + 1)(x - 4)(x^2 - 7)$ ?

1 Answer

Explanation:

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How do you find all zeros for (x+1)(x−4)(x2−7)(x + 1)(x - 4)(x^2 - 7)?

1 Answer

Explanation:

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How do you find all zeros for $(x + 1) (x - 4) (x^{2} - 7)$ $(x + 1)(x - 4)(x^2 - 7)$ ?