Question

Can someone please help me! I don’t understand how to do this!

Answer 1

A. 3 and 1/3 (Fraction), B. Has infinite answer (I think. I may have this wrong), and C. 0.

Explanation:

A. We have the equation 4(x-2)(4+6)= 4(x+2)^2 - 64.

1. (4x-8)(10)=(4x+8)^2 - 64 (We have to use PEMDAS, so the P goes first).

2.(4x-8)(10)= 16x +64 - 64 (Then we use E next.)

3. 40x-80 = 16x (Again with the P)

4. 40x = 16x +80 (Add 80 to both sides. And you might have drop the PEMDAS Here I think).

5. 24x = 80 (We subtract 16x from both sides)

6. x = 3 and 1/3 (Divide both sides by 24.

B. We have the equation: -3(x+2)(x-6)=-3(x-2)^2+48

1. Times the -3.

(-3x-6)(x-6)=(-3x-6)^2 +48

2. Times the exponent

(-3x-6)(x-6)=9x+36+48

3. Times the the ones that are in the braces the first using FOIL and add the other on the other side of the equation.

-3x^2 + 12x +36= 9x+84

4. Subtract 9x and 36 from both sides

-3x^2 +3x= 48

This may have an infinite answer to this, and I don't know if I'm wrong on this.

C. (x+5)(x+7)= (x +6)^2 -1

1. Times the exponent

(x+5)(x+7) = x^2 + 36 -1

2. Times the ones in the braces using FOIL.

x^2+7x+5x+35=X^2+36-1

3. Add all of the like Terms.

x^2+12x+35=x^2+35

4. Subtract x^2 from both sides (I may did this wrong)

12x + 35 = 35

5. Subtract 35 from both sides.

12x = 0

6. Divide both sides by 12 (I may have this wrong too)

x=0

And that's it. I may have some things wrong, but it's worth a shot.

Answer 2

9514 1404 393

Answer:

a. 4x^2 +16x -48 = 4x^2 +16x -48

b. -3x^2 +12x +36 = -3x^2 +12x +36

c. x^2 +12x +35 = x^2 +12x +35

Explanation:

In general, you would prove this by transforming one of the expressions into the other. The easiest would be to transform the factored form into the vertex form, perhaps, as this would spare you trying to explain the magic of factoring.

Alternatively, you can transform both expressions into the same (standard) form. I believe that will be the easiest of all.

The product of two binomials is ...

(x +a)(x +b) = x(x +b) +a(x +b) = x^2 +bx +ax +ab

(x +a)(x +b) = x^2 +(a+b)x +ab . . . . after collecting terms

The square of a binomial is the same thing, but with b=a, so ...

(x +a)^2 = x^2 +2a +a^2

Using these forms, we can avoid showing all of the intermediate "work" of making the desired transformations.

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a. 4(x -2)(x +6) = 4(x +2)^2 -64

4(x^2 +4x -12) = 4(x^2 +4x +4) -64 . . . . . expanding the products

4x^2 +16x -48 = 4x^2 +16x +16 -64 . . . . using the distributive property

4x^2 +16x -48 = 4x^2 +16x -48 . . . . . collect terms; expressions are equal

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b. -3(x +2)(x -6) = -3(x -2)^2 +48

-3(x^2 -4x -12) = -3(x^2 -4x +4) +48

-3x^2 +12x +36 = -3x^2 +12x -12 +48

-3x^2 +12x +36 = -3x^2 +12x +36

__

c. (x +5)(x +7) = (x +6)^2 -1

x^2 +12x +35 = x^2 +12x +36 -1

x^2 +12x +35 = x^2 +12x +35