Question

How do you write $y = 4{x^2} + 11x + 6$ in factored form?

Answer 1

Hint: Here, we are required to factorize a given expression using appropriate identities. First of all, we will identify the nature of the given expression whether it is a linear, quadratic, cubic or any other expression. Then, we will use the appropriate identity to break the expression in that form, and then finally, applying the identity will leave us with the required factors of the given expression.

Formula Used:
${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$

Complete step-by-step answer:
A quadratic equation is an equation having an integral non-negative power and it can always be represented in the form of $a{x^2} + bx + c$. Now, in this question, this given expression is known as a quadratic expression because it is of the form $a{x^2} + bx + c$ and it is having a power 2. Also, if we try to find the roots of this expression, then we will be able to find two roots only.
Given quadratic expression is:
$y = 4{x^2} + 11x + 6$
Now, we can write it in the form of:
$y = {\left( {2x} \right)^2} + 2\left( {2x} \right)\left( {\dfrac{{11}}{4}} \right) + {\left( {\dfrac{{11}}{4}} \right)^2} - {\left( {\dfrac{{11}}{4}} \right)^2} + 6$………………………………….$\left( 1 \right)$
Also, if we try to compare this with the given expression, then we can see that it is the same expression and only the way of writing it has changed.
Now, comparing the equation $\left( 1 \right)$ with:
${a^2} + 2ab + {b^2}$
We can use the appropriate identity:
${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$
Hence, comparing this with equation $\left( 1 \right)$ and substituting $2x = a$ and $\dfrac{{11}}{4} = b$ , we get,
$ \Rightarrow y = {\left( {2x + \dfrac{{11}}{4}} \right)^2} + 6 - \dfrac{{121}}{{16}} = {\left( {2x + \dfrac{{11}}{4}} \right)^2} + \dfrac{{96 - 121}}{{16}}$
$ \Rightarrow y = {\left( {2x + \dfrac{{11}}{4}} \right)^2} - \dfrac{{25}}{{16}}$
Now, rewriting this as:
$ \Rightarrow y = {\left( {2x + \dfrac{{11}}{4}} \right)^2} - {\left( {\dfrac{5}{4}} \right)^2}$
Using the identity ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$, we get,
$ \Rightarrow y = \left( {2x + \dfrac{{11}}{4} - \dfrac{5}{4}} \right)\left( {2x + \dfrac{{11}}{4} + \dfrac{5}{4}} \right)$
$ \Rightarrow y = \left( {2x + \dfrac{6}{4}} \right)\left( {2x + \dfrac{{16}}{4}} \right)$
Hence, taking LCM and solving further, we get,
$ \Rightarrow y = \left( {2x + \dfrac{3}{2}} \right)\left( {2x + 4} \right)$
$ \Rightarrow y = \left( {\dfrac{{4x + 3}}{2}} \right)\left( {2x + 4} \right)$
Taking 2 common from the second bracket,
$ \Rightarrow y = \left( {\dfrac{{4x + 3}}{2}} \right) \times 2 \times \left( {x + 2} \right)$
Cancelling out 2 from the numerator and denominator, we get,
$ \Rightarrow y = \left( {4x + 3} \right)\left( {x + 2} \right)$

Therefore, the factored form of $y = 4{x^2} + 11x + 6$ is $y = \left( {4x + 3} \right)\left( {x + 2} \right)$.
Hence, this is the required answer.

Note:
Alternatively, we can also solve this question using the middle term split as below:
Given expression is:
$y = 4{x^2} + 11x + 6$
Now, doing middle term split:
$ \Rightarrow y = 4{x^2} + 11x + 6 = 4{x^2} + 8x + 3x + 6$
\[ \Rightarrow y = 4x\left( {x + 2} \right) + 3\left( {x + 2} \right)\]
Now, taking the brackets common, we get
\[ \Rightarrow y = \left( {4x + 3} \right)\left( {x + 2} \right)\]
Therefore, we can also use middle term splitting for finding the factored form of a given quadratic equation.