how do you multiply these (x+h-c)(x-c)?

Answer 1

Simple distributive property! Like this:

(x+h-c)(x-c) =

(x*x + x*h + x*c) - (c*x + c*h - c*c) =

x^2 + hx + cx - cx - ch + c^2 =

x^2 + hx - ch + c^2

HTH! :-)

Answer 2

well first you would look for like terms, there are 2 x's and 2 c's. you would multiply the 2 x's and get 2x. then you would multiply the 2 c's and get 2c then you just add on the h. the answer is...

2x+2c+h

hope this helped

Answer 3

You apply the distributive property.

First you take the "x" in the first parentheses times each thing in the second parentheses: x*x and x(-c).
Then you distribute the "h" : h(x) and h(-c)
Finally you distribute the "-c": (-c)(x) and (-c)(-c).

So you have x^2 - cx + hx - ch - cx + c^2.

You can combine like terms and get:
x^2 - 2cx + hx - ch + c^2

Answer 4

it's the same as ones with just two.

(x-c)(x+h-c)
first take the first x, multiply it with the second x, the h, and the -c.
take the first -c, multiply it with the x, h, and -c.
so it should look like...
x(x)+x(h)+x(-c)+(-c)(x)+(-c)(h)+(-c)(-c)

then simplify

Answer 5

You would have to multiply out each term.

(x)(x) + x(-c) + h(x) + h(-c) + -c(x) + -c(-c) =

x^2 -cx +hx -hc -cx +c^2

x^2-2cx+hx-hc+c^2

Mysstere

Answer 6

x2+ xh-xc-cx-ch+c2=x2+hx-2cx-ch+c2