Question

What are the real zeros of the function g(x) = x3 + 2x2 − x − 2? 1, –1, 2 1, –1, –2 1, –1 2, –2, 1

Answer 1

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g(x) = x³ + 2x² - x - 2

To find the zeros or the roots of a function, you need to set the function equal to 0 then solve for x.

x³ + 2x² - x - 2 = 0

Factor left side of the equation

(x + 1)(x - 1)(x + 2) = 0

Now we can set factors equal to 0

x + 1= 0 or x - 1 = 0 or x + 2 = 0

x= 0 - 1 or x = 0 + 1 or x= 0 - 2

x= -1 or x = 1 or x = -2

Thus,

The roots or the zeros of the function are:

-2, -1, 1

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Answer 2

Answer:

Zeros are x=-2, -1, 1

Step-by-step explanation:

$g(x) = x^3 + 2x^2 - x -2$

We use grouping method

We group first two terms and last two terms

$g(x) = (x^3 + 2x^2)+(- x -2)$

Factor out GCF from each group

$g(x) = x^2(x+ 2)-1(x +2)$

FActor out x+2

$g(x) =(x+ 2)(x^2-1)$

To find out zeros , set g(x)=0

$0=(x+ 2)(x^2-1)$

Now we set each factor =0 and solve for x

$x+2=0$, x= -2

$x^2-1=0$

$x^2=1$, take square root on both sides

x=1, x=-1

Zeros are x=-2, -1, 1