+0  
 
0
339
2
avatar

Simplifying expression help please!!!, these are so hard

 Oct 31, 2017
 #1
avatar+7348 
+1

e.   To make factoring easier, rearrange the 25 - 20x + 4x2 .

 

     \(\frac{6x^2-11x-10}{6x^2-5x-6}\cdot\frac{6-4x}{4x^2-20x+25}\)

 

Now we need to factor all the numerators and denominators. Start by splitting their middle terms.

 

=   \(\frac{6x^2+4x-15x-10}{6x^2-9x+4x-6}\cdot\frac{6-4x}{4x^2-10x-10x+25}\)

                                                                            Now factor by grouping.

=   \(\frac{2x(3x+2)-5(3x+2)}{3x(2x-3)+2(2x-3)}\cdot\frac{6-4x}{2x(2x-5)-5(2x-5)}\)

 

=   \(\frac{(3x+2)(2x-5)}{(2x-3)(3x+2)}\cdot\frac{-2(2x-3)}{(2x-5)(2x-5)}\)

                                                                           Multiply the fractions together.

=   \(\frac{(3x+2)(2x-5)(-2)(2x-3)}{(2x-3)(3x+2)(2x-5)(2x-5)}\)

                                                                           Cancel the common terms.

=   \(\frac{{\color{red}(3x+2)}{\color{red}(2x-5)}(-2){\color{red}(2x-3)}}{{\color{red}(2x-3)(3x+2)(2x-5)}(2x-5)}\)

 

=   \(\frac{(-2)}{(2x-5)}\)

 

=   - \(\frac{2}{2x-5}\)

 

And  x  ≠  3/2  or  -2/3   since these cause a zero in the denominator of the original expression.

 Oct 31, 2017
 #2
avatar+7220 
+1

f)

\(\dfrac{3x^3-3a^2x}{x^2-2ax+a^2}\cdot\dfrac{a-x}{a^3x+a^2x^2}\\ =\dfrac{3x(x-a)(x+a)}{(x-a)(x-a)}\cdot\left(-\dfrac{x-a}{a^2x(a+x)}\right)\\ =\dfrac{-3x(x-a)^2(x+a)}{a^2x(x-a)^2(x+a)}\\ =-\dfrac{3}{a^2}\)

.
 Nov 1, 2017

17 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.