Question

find the real solutions 4(x+1)^2+23(x+1)+15=0

Answer 1

You can factor it as is, or... Make a substitution to make this easier.

Let u = x+1

then your equation is

4u^2 + 23u + 15 = 0

try factoring this.

Answer 2

u(4u+23)+-15 , U+ -15 or 2

Answer 3

mistake U= -15 or 2

Answer 4

That doesn't look factored right at all. You need to first multiply the first number by the last number,

4u^2 + 23u + 15 = 0

4*15=60.
Now find two numbers which multiply to make 60, and add up to the middle number, 23.

Answer 5

20 and 3

Answer 6

so now you have (4u+3)(u+5)

Answer 7

(4u+3)(u+5) = 0

now just replace u with x+1 and solve for x.

Answer 8

thanks, I got the answer

Answer 9

ttheres one more problem........solve by factoring 4x-11=3/x

Answer 10

arent i suppose to get rid of the x in the bottom?

Answer 11

Yes, multiply both sides by x.

Then subtract 3 on both sides, then factor it.

Answer 12

You'll be factoring

4x^2 - 11x - 3 = 0

Answer 13

So -1/4 and 3...your awesome, I truly appreciate it. I understand now