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lions [1.4K]
3 years ago
12

Select all the factors roots of x^3 +2x^2-16x-32

Mathematics
2 answers:
Ivanshal [37]3 years ago
5 0

Factor by grouping. Group up the terms into pairs, factor each pair, then factor out the overall GCF.

x^3 + 2x^2 - 16x - 32

(x^3 + 2x^2) + (-16x-32) ... pair up terms

x^2(x + 2) + (-16x - 32) ... factor x^2 from the first group

x^2(x + 2) - 16(x + 2) ... factor -16 from the second group

(x^2 - 16)(x + 2) .... factor out (x+2)

(x - 4)(x + 4)(x + 2) .... Use the difference of squares to factor x^2-16

---------------------------

The original expression completely factors to (x - 4)(x + 4)(x + 2)

The three factors are x - 4  and  x + 4   and   x + 2

solmaris [256]3 years ago
5 0

The answers are b) -2, c) -4, and e) 4.


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I need help seeing what I did wrong
Makovka662 [10]

Answer:  The two roots are x = 3/2 and x = -2

=========================================================

Explanation:

You have the right idea so far. But the two numbers should be 3 and -4 since

  • 3*(-4) = -12
  • 3+(-4) = -1

The -1 being the coefficient of the x term.

This means you need to change the -3x and 4x to 3x and -4x respectively. The other inner boxes are correct.

---------

Refer to the diagram below to see one way to fill out the box method, and that helps determine the factorization.

If we place a 2x to the left of -2x^2, then we need an -x up top because 2x*(-x) = -2x^2

Then based on that outer 2x, we need a -2 up top over the -4x. That way we get 2x*(-2) = -4x

So we have the factor -x-2 along the top

The last thing missing is the -3 to the left of 3x. Note how -3*(-x) = 3x in the left corner and -3*(-2) = 6 in the lower right corner.

We have the factor 2x-3 along the left side.

---------

The two factors are (2x-3) and (-x-2) which leads to the factorization (x+3)(-x+2)

The last thing to do is set each factor equal to 0 and solve for x

  • 2x-3 = 0 solves to x = 3/2 = 1.5
  • -x-2 = 0 solves to x = -2

The two roots are x = 3/2 and x = -2

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Answer:

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Step-by-step explanation:

If the scale of the map is 1in:25mi, and you measure 6.5in between two twoms, then the real distance would be 6.5in*25mi = 162.5mi.

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