How do you factor \[6{{x}^{2}}-11x+4?\]
Answer
Verified
442.2k+ views
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.
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.
Recently Updated Pages
What percentage of the area in India is covered by class 10 social science CBSE
The area of a 6m wide road outside a garden in all class 10 maths CBSE
What is the electric flux through a cube of side 1 class 10 physics CBSE
If one root of x2 x k 0 maybe the square of the other class 10 maths CBSE
The radius and height of a cylinder are in the ratio class 10 maths CBSE
An almirah is sold for 5400 Rs after allowing a discount class 10 maths CBSE
Trending doubts
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Why is there a time difference of about 5 hours between class 10 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE
Explain the Treaty of Vienna of 1815 class 10 social science CBSE
Write an application to the principal requesting five class 10 english CBSE