I dont know how to solve this equation,someone can help me?

Question

(x+4)(x-7) & 10x^2y^3+15x^2y^2-5xy

Answer 1

sry school doesnt start for 2 weeks so until then i refuse.

Answer 2

these are unsolvable. the first one can be distributed and the second one can be factored, but it is impossible to give numbers for the variables

1. (x+4)(x-7)

x^2+4x-7x-28

x^2-3x-28

2. 10x^2y^3+15x^2y^2-5xy

5xy(2xy^2+3xy-1)

make it a good day

Answer 3

0 = (x + 4)(x - 7) [If that is what you mean - solve that quadratic when it equals 0]

0 = x + 4

=> x = -4, OR

0 = x - 7

=> x = 7

(x + 4)(x - 7) [If you mean to expand]

= x^2 + 4x - 7x - 28

=> = x^2 - 3x - 28

I am not skilled enough to do the other one. =(

Answer 4

Lets go for the first one

Assuming you meant 0=x(x-7)+4(x-7)

then First lets expand it

0=x^2-7x+4x-28

then take the absolute number on the right (28) by adding the number on both sides

0+28=x^2-7x+40x-28+28

to get

28=x^2+3x

now we solve for x (brain lock can someone pick it up from here) i can see the answer is x=4 or x=-7 but can't just seem to decipher it