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Find all the zeros of f(x).

f(x) = 2x3 + 7x2 - 28x + 12
Arrange your answers from smallest to largest. If
there is a double root, list it twice.

Plz help!

Find all the zeros of f(x). f(x) = 2x3 + 7x2 - 28x + 12 Arrange your answers from-example-1
User Igauravsehrawat
by
2.6k points

2 Answers

15 votes
15 votes

Answer: -6, 1/2, 2.

Explanation:

{(x) = 2x3 + 7x2 - 28x + 12 = 0

From the first and last coefficient 2 and 12, one

guess for a zero is x=2.

So substituting x=2:

{(2) = 16+ 28 - 56 + 12 = 0

So x = 2 is a zero and x - 2 is a factor of f(x)

Performing long division:

X - 2)2x3 + 7x2 - 28x + 12(2x2 + 11x - 6 <

Quotient

User Skaak
by
3.5k points
15 votes
15 votes

Answer:

The zeroes are -6, 1/2 and 2.

Explanation:

f(x) = 2x3 + 7x2 - 28x + 12 = 0

From the first and last coefficient 2 and 12, one guess for a zero is x = 2.

So substituting x = 2:

f(2) = 16+ 28 - 56 + 12 = 0

So x = 2 is a zero and x - 2 is a factor of f(x)

Performing long division:

x - 2)2x3 + 7x2 - 28x + 12( 2x2 + 11x - 6 <------- Quotient

2x3 - 4x2

11x2 - 28x

11x2 - 22x

- 6x + 12

-6x + 12

.............

Now we solve

2x2 + 11x - 6 = 0

(2x - 1)(x + 6) = 0

2x - 1 = 0 or x + 6 = 0, so:

x = 1/2, x = -6.