Trying to help daughter solve this easy one.....?

You need to multiply each term to the other

(3x)(2x) + (3x)(7) - (4) (2x) - (4)(7)

6x^2 + 21x - 8x - 28

6x^2 + 13x - 28

Good luck

There is something missing from the problem as stated.

Did you mean

(1) Solve (3x-4)(2x+7) = 0

or

(2) Carry out the multiplication (3x-4)(2x+7).

For (1):

(3x-4)(2x+7) = 0

The only way two factors can multiply and give 0 is if one of the two factors is equal to 0.

So either 3x-4 = 0 or 2x+7 = 0.

In case 3x-4 = 0 then we get 3x = 4, and then x = 3/4.

In case 2x+7 = 0 then we get 2x = -7, and then x = -7/2.

Thus there are two possible values for x, 3/4 and -7/2.

For (2):

Using the FOIL method we get:

6x^2 + 21x - 8x - 28,

which simplifies to

6x^2 + 13x - 28.

(NOTE: on the keyboard, we use x^2 to mean x to the 2nd power, or x-squared.)

Maybe one of these will help.

If you are solving for x then you get two answers. You must first break the problem into:

3x-4=0

and

2x+7=0

You then solve for X giving you 4/3 and -7/2 respectively.

It would help to know the context of the question.

Good luck!

first multiply 3x by 2x, then multiply 3x by 7 and add them.

3x*2x + 3x*7 ---> 6x^2 + 21x

then add to that -4 times 2x, then -4 times 7

6x^2 + 21x - 4*2x - 4*7

6x^2 + 21x - 8x - 28 ---> 6x^2 + 13x - 28

to solve those you "FOIL" (First outside, Inside Last)

in variable form, you solve like this:

if you start with (a+b)(c+d)

then first multiply a by both things in the 2nd set of parenthases (c and d), then add those two products together (because c and d are added together)

so you'll have: ac + ad

then multiply b by both things in the 2nd set of parentheses (c and d), then add those two products to what you already had (ac + ad)

so you'll have ac + ad + bc + bd

make sure that if there's anything that can be made simpler, you do it (e.g. if you end up with x^2 + 4x + 7x +2, then make sure you simplify it to x^2 + 11x + 2)

(you can also research the Distributive Property online, because that's what that's called)

(3x-4)(2x+7)

Use FOIL to get next step (first, outside, inside, last)

6x^2 + 21x - 8x - 28

Next, add like terms

6x^2 + 13x - 28

Next, you want to use the quadratic equation which is:

See : http://mathworld.wolfram.com/images/equations/Quad...

You will use A = 6, B=13, and C= -18

3x-4 is -12. 2x+7 is the same as 2x7 which is 14. and the way this is set out is all wrong it should say (3x-4)+(2x7) which is 2. because a negative number plus a positive number means that the negative goes up. eg -4+7=3.

FOIL (first, outer, inner, last)

(3x-4)(2x+7)

6x^2 + 21x - 6x -28

F 0 I L

6x^2-13x-28 is the answer! Good luck!

(3x-4)=0 (2x+7)=0

3x=4 2x=-7

x=4/3 or x=-7/2

is the qns incomplete?

u wanna solve x?

all i got was

6x(squared) + 21x - 8 x - 21= 6x(squared) +13x -21...

loren, clooney and happy are the only ones to get this right.