Answer

Verified

419.7k+ views

**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$

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Trending doubts

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Which are the Top 10 Largest Countries of the World?

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

How do you graph the function fx 4x class 9 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Change the following sentences into negative and interrogative class 10 english CBSE