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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.

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}$

$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

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$