
Factorize $3{x^2} = x + 4.$
Answer
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Hint:Factorization is the process of breaking the polynomial entity into its factors. One such method for factorization is the Grouping method. Grouping method: In this method we work upon the expression which is of this form ${x^2} + ax + b$and then look for two factors $p\;\& \;q$such that $p \times q = b\,\;\& \;p + q = a$
So here for the given polynomial we can perform factorization by grouping method and thus find its factors.
Complete step by step solution:
Given
$3{x^2} = x + 4.........................\left( i \right)$
Now we can write (i) in the standard form as:
$
\Rightarrow 3{x^2} = x + 4 = 3{x^2} - x - 4 = 0 \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3\left( {{x^2} - \left( {\dfrac{x}{3}} \right) -
\left( {\dfrac{4}{3}} \right)} \right)...................\left( {ii} \right) \\
$
Now in order to factorize it we use the Grouping method.
Grouping method: In this method we work upon the expression which is of this form ${x^2} + ax + b$and then look for two factors $p\;\& \;q$such that
$p \times q = b\,\;\& \;p + q = a....................\left( {iii} \right)$
Then the given expression can be written as
${x^2} + (a + b)x + ab$
Multiply $x$inside$\left( {p + q} \right)$, we get:
These expressions that we get, $(x + a)\;,(x + b)$ are the factors of the given algebraic expression
Now on comparing with (ii) and (iii) we can write $p = - 4$and$q = 3$. Thus:
$
- 4 \times 3 = - 12 = b \\
\left( { - 4} \right) + 3 = - 1 = a \\
$
Now on using Grouping Method we can write:
\[
3{x^2} - x - 4 \\
= 3\left( {{x^2} - \left( {\dfrac{x}{3}} \right) - \left( {\dfrac{4}{3}} \right)} \right) \\
= 3\left( {{x^2} + \left( {\dfrac{{ - 4 + 3}}{3}} \right)x + \left( {\dfrac{{ - 4 \times 3}}{3}} \right)} \right)
\\
= 3\left( {{x^2} + \left( {\dfrac{{ - 4}}{3}} \right)x + \left( {\dfrac{3}{3}} \right)x + \left( {\dfrac{{ -
12}}{3}} \right)} \right) \\
= \left( {3{x^2} - 4x + x + \left( {\dfrac{{ - 12}}{3}} \right)} \right) \\
= \left( {3{x^2} - 4x + x + - 4} \right).................................\left( {iv} \right) \\
\]
Now we have to take the common terms from each of the terms.
$
= x(3x - 4) + 1(x - 4) \\
= (3x - 4)(x + 1) \\
$
Then on factoring ${x^2} + 4x - 21$we get the factors as $(3x - 4)\,\;and\;\;(x + 1)$ .
Note: The Grouping method is used when the algebraic expression given is in the form where the 2 nd term and 3 rd can be split into products and sums as shown above.
If the given algebraic expression is in the form of any known identity then we can perform factorization easily using conventional methods or else methods like grouping are used. In grouping methods care should be taken while choosing $p\;and\;q$since it is an area where human errors can interfere.
So here for the given polynomial we can perform factorization by grouping method and thus find its factors.
Complete step by step solution:
Given
$3{x^2} = x + 4.........................\left( i \right)$
Now we can write (i) in the standard form as:
$
\Rightarrow 3{x^2} = x + 4 = 3{x^2} - x - 4 = 0 \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3\left( {{x^2} - \left( {\dfrac{x}{3}} \right) -
\left( {\dfrac{4}{3}} \right)} \right)...................\left( {ii} \right) \\
$
Now in order to factorize it we use the Grouping method.
Grouping method: In this method we work upon the expression which is of this form ${x^2} + ax + b$and then look for two factors $p\;\& \;q$such that
$p \times q = b\,\;\& \;p + q = a....................\left( {iii} \right)$
Then the given expression can be written as
${x^2} + (a + b)x + ab$
Multiply $x$inside$\left( {p + q} \right)$, we get:
These expressions that we get, $(x + a)\;,(x + b)$ are the factors of the given algebraic expression
Now on comparing with (ii) and (iii) we can write $p = - 4$and$q = 3$. Thus:
$
- 4 \times 3 = - 12 = b \\
\left( { - 4} \right) + 3 = - 1 = a \\
$
Now on using Grouping Method we can write:
\[
3{x^2} - x - 4 \\
= 3\left( {{x^2} - \left( {\dfrac{x}{3}} \right) - \left( {\dfrac{4}{3}} \right)} \right) \\
= 3\left( {{x^2} + \left( {\dfrac{{ - 4 + 3}}{3}} \right)x + \left( {\dfrac{{ - 4 \times 3}}{3}} \right)} \right)
\\
= 3\left( {{x^2} + \left( {\dfrac{{ - 4}}{3}} \right)x + \left( {\dfrac{3}{3}} \right)x + \left( {\dfrac{{ -
12}}{3}} \right)} \right) \\
= \left( {3{x^2} - 4x + x + \left( {\dfrac{{ - 12}}{3}} \right)} \right) \\
= \left( {3{x^2} - 4x + x + - 4} \right).................................\left( {iv} \right) \\
\]
Now we have to take the common terms from each of the terms.
$
= x(3x - 4) + 1(x - 4) \\
= (3x - 4)(x + 1) \\
$
Then on factoring ${x^2} + 4x - 21$we get the factors as $(3x - 4)\,\;and\;\;(x + 1)$ .
Note: The Grouping method is used when the algebraic expression given is in the form where the 2 nd term and 3 rd can be split into products and sums as shown above.
If the given algebraic expression is in the form of any known identity then we can perform factorization easily using conventional methods or else methods like grouping are used. In grouping methods care should be taken while choosing $p\;and\;q$since it is an area where human errors can interfere.
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