Find P(7).................

Is X +3 a factor of 2X squared minus 2X -12

Fantom [35]

Answer: No, x+3 is not a factor of 2x^2-2x-12

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Explanation:

Let p(x) = 2x^2 - 2x - 12

If we divide p(x) over (x-k), then the remainder is p(k). I'm using the remainder theorem. A special case of the remainder theorem is that if p(k) = 0, then x-k is a factor of p(x).

Compare x+3 = x-(-3) to x-k to find that k = -3.

Plug x = -3 into the function

p(x) = 2x^2 - 2x - 12

p(-3) = 2(-3)^2 - 2(-3) - 12

p(-3) = 12

We don't get 0 as a result so x+3 is not a factor of p(x) = 2x^2 - 2x - 12

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Let's see what happens when we factor p(x)

2x^2 - 2x - 12

2(x^2 - x - 6)

2(x - 3)(x + 2)

The factors here are 2, x-3 and x+2

3 0

2 years ago

Vedanta Excel In Opt. Mathematics - Book 9

givi [52]

<h2> Distance formual</h2>

<h2>= √(2-1)^+(-3-2)^</h2><h2>= √(1)^+(-5)^ [here square root and </h2><h2> square get cancelled]</h2><h2>= (1)+(-5)</h2><h2>= 1-5</h2><h2>= - 4</h2>

6 0

3 years ago

When translating a figure using a combination of two translations, is the resulting figure congruent to the original figure?

emmainna [20.7K]

It has to be the original figure but it is in a different spot on the graph

6 0

3 years ago

What is the quotient? 2m+4/8 m+2/6

enot [183]

(2m + 4) / 8 x 6/m+2

2(m + 2)/8 x 6/ (m+2) : 2 from [2(m] becomes a 1. 6 becomes a 3 , 8 becomes a 4 then cross out 4 and it becomes a 2.

= 3/2

5 0

3 years ago

Read 2 more answers

PLEASE HELP

SVETLANKA909090 [29]

Answer:

Hill 1: F(x) = -(x + 4)(x + 3)(x + 1)(x - 1)(x - 3)(x - 4)

Hill 2: F(x) = -(x + 4)(x + 3)(x + 1)(x - 1)(x - 3)(x - 4)

Hill 3: F(x) = 4(x - 2)(x + 5)

Step-by-step explanation:

Hill 1

You must go up and down to make a peak, so your function must cross the x-axis six times. You need six zeros.

Also, the end behaviour must have F(x) ⟶ -∞ as x ⟶ -∞ and F(x) ⟶ -∞ as x⟶ ∞. You need a negative sign in front of the binomials.

One possibility is

F(x) = -(x + 4)(x + 3)(x + 1)(x - 1)(x - 3)(x - 4)

Hill 2

Multiplying the polynomial by -½ makes the slopes shallower. You must multiply by -2 to make them steeper. Of course, flipping the hills converts them into valleys.

Adding 3 to a function shifts it up three units. To shift it three units to the right, you must subtract 3 from each value of x.

The transformed function should be

F(x) = -2(x +1)(x)(x -2)(x -3)(x - 6)(x - 7)

Hill 3

To make a shallow parabola, you must divide it by a number. The factor should be ¼, not 4.

The zeroes of your picture run from -4 to +7.

One of the zeros of your parabola is +5 (2 less than 7).

Rather than put the other zero at ½, I would put it at (2 more than -4) to make the parabola cover the picture more evenly.

The function could be

F(x) = ¼(x - 2)(x + 5).

In the image below, Hill 1 is red, Hill 2 is blue, and Hill 3 is the shallow black parabola.

6 0

3 years ago