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Kamila [148]
3 years ago
15

Zeros of polynomials (with factoring

Mathematics
2 answers:
Anon25 [30]3 years ago
3 0

Answer:

x = -5/2    x=1      x = -1

Step-by-step explanation:

p(x) = 2x^3 + 5x^2 – 2x – 5

Use factor by grouping

p(x) = 2x^3 + 5x^2       – 2x – 5

Factor x^2 from the first group and -1 from the second group

    x^2(2x +5) -1( 2x+5)

Then factor out 2x+5

p(x) = (2x+5) (x^2-1)

Factor x^2 -1 as the difference of squares

p(x) = (2x+5)(x-1)(x+1)

Set to zero to find the x intercepts

0 = (2x+5)(x-1)(x+1)

Using the zero product property

2x+5 =0    x-1 =0   x+1 =0

2x = -5        x=1       x=-1

x = -5/2    x=1      x = -1

kirill115 [55]3 years ago
3 0

Answer:

The zeroes are (-1,0), (1, 0) and (-5/2, 0)

Step-by-step explanation:

We can find the zeroes by factoring:

2x^3 + 5x^2 - 2x - 5 = 0  

x^2(2x + 5) - 1(2x + 5) = 0

(x^2 - 1)(2x + 5) = 0

(x - 1)(x + 1)(2x + 5) = 0

So x = -1, 1, -5/2.

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Answer:

\mu = 47.17

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Step-by-step explanation:

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3 years ago
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