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# How do you factorize the expression $6{x^2} - 33x + 15$?  Verified
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Hint: In this question, we are given a quadratic expression and we have to factorize it. Use splitting the middle term method to find the factors. Let us assume that the equation is in the form of $a{x^2} + bx + c$.
If ${x^2}$ does not have a coefficient, then find two factors of $c$, such that when they are added or subtracted, they give us $b$.
But, if ${x^2}$ has a coefficient, then find two factors of $ac$, such that when they are added or subtracted, they give us $b$. After finding factors, take out the common numbers or variables and factorize the expression.

Complete step-by-step solution:
We are given a quadratic expression $6{x^2} - 33x + 15$.
Let us find two such factors of $90{\text{ }}( = 6 \times 15)$, such that when they are added or subtracted, they give us $- 33$.
If we observe, then two such factors are $30$ and $3$.
But, if we add them, we will get $33$ and we want $- 33$.
So, we will take our factors in the negative. Our required factors are $- 30$ and $- 3$.
$\Rightarrow 6{x^2} - 3x - 30x + 15$
Taking $3x$ common from the first two terms and $15$ from the last two terms,
$\Rightarrow 3x\left( {2x - 1} \right) - 15\left( {2x - 1} \right)$
Making factors,
$\Rightarrow \left( {3x - 15} \right)\left( {2x - 1} \right)$
Hence, these are the factors of $6{x^2} - 33x + 15$.
Using the factors, let us find the values of $x$.
$\Rightarrow \left( {3x - 15} \right)\left( {2x - 1} \right)$
Let us equate to zero,
$\Rightarrow 3x - 15 = 0,2x - 1 = 0$
Shifting to find the required values,
$\Rightarrow 3x = 15,2x = 1$
$\Rightarrow x = 5,\dfrac{1}{2}$

Hence, the values of $x$ are equal to $5$ and $\dfrac{1}{2}$

Note: We can also factorize the expression using the formula –
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
But using this formula, we will directly get the values of $x$. Using the values, we will have to find the factors. Let us see how.
We know that $a = 6,b = - 33,c = 15$. Let us put the values in the formula –
$\Rightarrow x = \dfrac{{33 \pm \sqrt {{{33}^2} - 4 \times 6 \times 15} }}{{2 \times 6}}$
Simplifying,
$\Rightarrow x = \dfrac{{33 \pm \sqrt {1089 - 360} }}{{12}}$
Let us subtract we get
$\Rightarrow x = \dfrac{{33 \pm \sqrt {729} }}{{12}}$
Now, we know that$\sqrt {729} = 27$.
$\Rightarrow x = \dfrac{{33 \pm 27}}{{12}}$
We will get –
$\Rightarrow x = \dfrac{{60}}{{12}},\dfrac{6}{{12}}$
$\Rightarrow x = 5,\dfrac{1}{2}$
Now, we will make factors using these values.
$\Rightarrow x - 5 = 0,x - \dfrac{1}{2} = 0$
$\Rightarrow x - 5 = 0,2x - 1 = 0$
Hence, the two factors are $\Rightarrow \left( {x - 5} \right)\left( {2x - 1} \right) = 0$