How do you multiply #\frac { x - 4} { 4x ^ { 2} + 10x } \cdot \frac { 4x ^ { 3} + 10x ^ { 2} } { 2x ^ { 2} }#?

1 Answer
Apr 24, 2017

#(x-4)/(2x)#

Explanation:

When multiplying fractions we simply multiply across.
Let's start with the numerators.

#(x-4)(4x^3+10x^2)#

Remember we multiply every term with each other.
So,

#(x*4x^3)+(x*10x^2)+(-4*4x^3)+(-4*10x^2)#

#4x^4+10x^3-16x^3-40x^2#

[Simplify common variables]

#4x^4color(red)(+10x^3-16x^3)-40x^2#

#4x^4-6x^3-40x^2#

#color(blue)"Now we do the same for the denominators"#

#(4x^2+10x)(2x^2)#

#(4x^2*2x^2)+(10x*2x^2)#

#8x^4+20x^3#

So now we have...
our numerator
#4x^4-6x^3-40x^2#

and our denominator
#8x^4+20x^3#

#(4x^4-6x^3-40x^2)/(8x^4+20x^3)#

Now look for common factors to simplify our fraction.
I found #2x^2# in the numerator and #4x^3# in the denominator

#(2x^2(2x^2-3x-20))/(4x^3(2x+5))#

Divide our common factor

#((2x^2-3x-20))/(2x(2x+5))#

#(2x^2-3x-20)/(2x(2x+5)#

Factor the numerator using whatever method your prefer.
Look here if you are confused on how I factored the numerator.
Your result should be

#((x-4)(2x+5))/(2x(2x+5)#

#((x-4)cancel((2x+5)))/(2xcancel((2x+5))#

#(x-4)/(2x)#