Solveeit Logo
Login

Question

Question: How do you factor \(6{{x}^{2}}+7x-3\)?...

How do you factor 6x2+7x−36{{x}^{2}}+7x-3?

Explanation

Solution

We use both grouping method and vanishing method to solve the problem. The quadratic equation is 6x2+7x−36{{x}^{2}}+7x-3. We take common terms out to form the multiplied form of different polynomials. In the case of vanishing method, we use the value of xx which gives the polynomial value 0.

Complete step by step solution:
We apply the middle-term factoring or grouping to factorise the polynomial.

Factorising a polynomial by grouping is to find the pairs which on taking their common divisor out,
gives the same remaining number.

In the case of 6x2+7x−36{{x}^{2}}+7x-3, we break the middle term 7x7x into two parts of 9x9x and −2x-2x.

So, 6x2+7x−3=6x2+9x−2x−36{{x}^{2}}+7x-3=6{{x}^{2}}+9x-2x-3. We have one condition to check if the grouping is possible
or not. If we order the individual elements of the polynomial according to their power of variable,
then the multiple of end terms will be equal to the multiple of middle terms.

Here multiplication for both cases gives −18x2-18{{x}^{2}}. The grouping will be done for 6x2+9x6{{x}^{2}}+9x
and −2x−3-2x-3.

We try to take the common numbers out.

For 6x2+9x6{{x}^{2}}+9x, we take 3x3x and get 3x(2x+3)3x\left( 2x+3 \right).

For −2x−3-2x-3, we take −1-1 and get −1(2x+3)-1\left( 2x+3 \right).

The equation becomes 6x2+7x−3=6x2+9x−2x−3=3x(2x+3)−1(2x+3)6{{x}^{2}}+7x-3=6{{x}^{2}}+9x-2x-3=3x\left( 2x+3 \right)-1\left( 2x+3 \right).

Both the terms have (2x+3)\left( 2x+3 \right) in common. We take that term again and get

& 6{{x}^{2}}+7x-3 \\\ & =3x\left( 2x+3 \right)-1\left( 2x+3 \right) \\\ & =\left( 2x+3 \right)\left( 3x-1 \right) \\\ \end{aligned}$$ **Therefore, the factorisation of $6{{x}^{2}}+7x-3$ is $$\left( 2x+3 \right)\left( 3x-1 \right)$$.** **Note:** We find the value of $x$ for which the function $f\left( x \right)=6{{x}^{2}}+7x-3=0$. We can see $f\left( \dfrac{1}{3} \right)=6{{\left( \dfrac{1}{3} \right)}^{2}}+7\left( \dfrac{1}{3} \right)- 3=\dfrac{2}{3}+\dfrac{7}{3}-3=0$. So, the root of the $f\left( x \right)=6{{x}^{2}}+7x-3$ will be the function $\left( x-\dfrac{1}{3} \right)$. This means for $x=a$, if $f\left( a \right)=0$ then $\left( x-a \right)$ is a root of $f\left( x \right)$. Now, $f\left( x \right)=6{{x}^{2}}+7x-3=\left( 2x+3 \right)\left( 3x-1 \right)$. We can also do the same process for $$\left( 2x+3 \right)$$.