Question

Factorise:
10(3x+1/x)^2-(3x+1/x)-3
Please help guys

Answer 1

ARYAN.....PVT.....LTD.....

Answer 2

Answer:

Factorized form of $10(3x+\frac{1}{x})^2-(3x+\frac{1}{x})-3$ is $(\frac{15x^2+5-3x}{x} )(\frac{6x^2+2+x}{x} )$

Step-by-step explanation:

Consider the given quadratic equation,

$10(3x+\frac{1}{x})^2-(3x+\frac{1}{x})-3$

We have to factorized the given equation.

Substitute $(3x+\frac{1}{x})=t$ then above quadratic equation becomes,

$10(3x+\frac{1}{x})^2-(3x+\frac{1}{x})-3$

$10t^2-t-3$

Using middle term splitting method,

$10\times(-3)=(-30)$ can be split as 5 and -6 as their sum is -1 and product is (-30).

$10t^2+5t-6t-3$

$5t(2t+1)-3(2t+1)$

$(5t-3)(2t+1)$

Putting value of $t=(3x+\frac{1}{x})$

$(5(3x+\frac{1}{x})-3)(2(3x+\frac{1}{x})+1)$

$(\frac{15x^2+5-3x}{x} )(\frac{6x^2+2+x}{x} )$

Thus, factorized form of $10(3x+\frac{1}{x})^2-(3x+\frac{1}{x})-3$ is $(\frac{15x^2+5-3x}{x} )(\frac{6x^2+2+x}{x} )$