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# How do you factor $6{{x}^{2}}-11x+4?$

Last updated date: 13th Jun 2024
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Hint: We are given $6{{x}^{2}}-11x+4,$ and ,we are asked to find the factor form of this. To do so we will first understand the type of equation we have, once we get that we will find the greatest common factor from each term then in the remaining term be factor using the middle term. We use $a\times b$ in such a way that its sum or difference from the ‘b’ of the equation $a{{x}^{2}}+bx+c.$ Once we split that we will unite all factors of the equation and we get our required answer.

Complete step-by-step solution:
We are given $6{{x}^{2}}-11x+4$ and we are asked to find the factor of it. To find the factor of the equation, we should see that as the highest power is 2 so it is 2 degree polynomial. So it is a quadratic equation. Now, to factor, we will first find the possible greatest common factor of all these. In 6, – 11 and 4, we can see that there is no common term to each of these terms. So we cannot separate anything out of this. Now we will use the middle term to split. In middle term split apply on $a{{x}^{2}}+bx+c,$ we produce ‘a’ by ‘c’ and then factor ‘ac’ in such a way that if the product is ‘ac’ while the sum or difference is made up to ‘b’. Now we have a middle term on ${{x}^{2}}+13x+12.$ We have a = 5, b = – 11 and c = 4. So, we use these values to find two terms which help us in splitting the middle term. Now we can see that,
$a\times c=6\times 4=24$
We can see that there are two terms – 8 and – 3 such that $-8\times -3=24\left[ \text{same as }a\times c \right]$ and $-8+\left( -3 \right)=-8-3=-11.$ So, we use this to split the middle term. So $6{{x}^{2}}-11x+4$ becomes
$\Rightarrow 6{{x}^{2}}+\left( -8-3 \right)x+4$
Opening the brackets
$\Rightarrow 6{{x}^{2}}-8x-3x+4$
We take common in the first 2 terms and the last 2 terms. So we get
$\Rightarrow 2x\left( 3x-4 \right)-1\left( 3x-4 \right)$
As (3x – 4) is the same, so we get,
$\Rightarrow \left( 2x-1 \right)\left( 3x-4 \right)$
So, we get,
$\Rightarrow 6{{x}^{2}}-11x+4=\left( 2x-1 \right)\left( 3x-4 \right)$
So, the factored form of $6{{x}^{2}}-11x+4$ is $\left( 2x-1 \right)\left( 3x-4 \right).$

Note: While finding the middle term using a factor of $a\times c,$ we need to keep in mind that when the sign of ‘a’ and ‘c’ are the same then ‘b’ is obtained by addition only, if the sign of ‘a’ and ‘c’ are different then ‘b’ can be obtained using only subtraction. So, as we have a = 6 and c = 4 have the same sign so ‘b’ is obtained as – 8 + (– 3) by addition of – 8 and – 3. We can always cross-check that. Product of $\left( 2x-1 \right)\left( 3x-4 \right)=2x\left( 3x-4 \right)-1\left( 3x-4 \right)$ on simplifying we get $6{{x}^{2}}-8x-3x+4.$ Adding the like terms, we get $6{{x}^{2}}-11x+4.$ So, our factors are correct.