Respuesta :
Answer:
Cancelling x and (2x-3), we get, the simplest form as
[tex]\frac{4x^3-10x^2+6x}{2x^3+x^2-3x}=\frac{2(x-1)}{(x+1)}[/tex]
Step-by-step explanation:
Consider the given two expressions [tex]4x^3-10x^2+6x[/tex] and [tex]2x^3+x^2-3x[/tex].
We solve both expressions seperately,
Consider the first expression [tex]4x^3-10x^2+6x[/tex]
Taking x common from the expression,
[tex]4x^3-10x^2+6x=x(4x^2-10x+6)[/tex]
The terms in brackets is a quadratic equation, we can solve using middle term splitting method,
[tex]4x^2-10x+6[/tex]
-10x can be written as -4x-6x , we get,
[tex]4x^2-10x+6=4x^2-4x-6x+6[/tex]
[tex]\Rightarrow 4x(x-1)-6(x-1)=(4x-6)(x-1)[/tex]
[tex]4x^3-10x+6)=2x(2x-3)(x-1)[/tex]
Consider the second term , [tex]2x^3+x^2-3x[/tex]
Taking x common from the expression we have,
[tex]2x^3+x^2-3x=x(2x^2+x-3)[/tex]
The terms in brackets is a quadratic equation, we can solve using middle term splitting method,
[tex]2x^2+x-3[/tex]
x can be written as 3x-2x
[tex]2x^2+x-3=2x^2+2x-3x-3[/tex]
[tex]2x^2+2x-3x-3=2x(x+1)-3(x+1)=(2x-3)(x+1)[/tex]
Thus, [tex]2x^3+x^2-3x=x(2x-3)(x+1)[/tex]
Our expression is [tex]4x^3-10x^2+6x[/tex] over [tex]2x^3+x^2-3x[/tex]
is [tex]\frac{4x^3-10x^2+6x}{2x^3+x^2-3x}[/tex]
[tex]\frac{4x^3-10x^2+6x}{2x^3+x^2-3x}=\frac{2x(2x-3)(x-1)}{x(2x-3)(x+1)}[/tex]
Cancelling same terms from numerator and denominator , thus cancelling x and (2x-3), we get, the simplest form as
[tex]\frac{4x^3-10x^2+6x}{2x^3+x^2-3x}=\frac{2(x-1)}{(x+1)}[/tex]
