Answer

Verified

419.4k+ views

**Hint:**Here, we have to factorise $12{\left( {{x^2} + 7x} \right)^2} - 8\left( {{x^2} + 7x} \right)\left( {2x - 1} \right) - 15{\left( {2x - 1} \right)^2}$ which is a biquadratic expression. First of all we suppose $\left( {{x^2} + 7x} \right) = a$ and $\left( {2x - 1} \right) = b$, then put these values in above expression and factorise it. After that put the value of $a$ and $b$ in factored terms and, we will get quadratic so again factorise it to reach a final answer.

**Complete step-by-step solution:**

Here, the given expression is $12{\left( {{x^2} + 7x} \right)^2} - 8\left( {{x^2} + 7x} \right)\left( {2x - 1} \right) - 15{\left( {2x - 1} \right)^2}$.

Suppose $\left( {{x^2} + 7x} \right) = a$and $\left( {2x - 1} \right) = b$, putting these values in above given expression.

We can write, $12{a^2} - 8ab - 15{b^2}$.

Now, to factorise the expression, break $ - 8ab$ in two parts such that their product is $ - 180{a^2}{b^2}$.

We can write $ - 8ab$ as $ - 18ab + 10ab$because its product is $ - 180{a^2}{b^2}$.

So, we can write expression as,

$ = 12{a^2} - 18ab + 10ab - 15{b^2}$

Now, take $6a$ as common from the first two terms and $5b$ as common from the last two terms. So, after applying this we can write,

$ = 6a\left( {2a - 3b} \right) + 5b\left( {2a - 3b} \right)$

Now, taking $\left( {2a - 3b} \right)$ common from both terms, we can write

$ = \left( {2a - 3b} \right)\left( {6a + 5b} \right)$

Putting the value of $a$ and $b$ in above factorisation, we can write

$ = \left( {2\left( {{x^2} + 7x} \right) - 3\left( {2x - 1} \right)} \right)\left( {6\left( {{x^2} + 7x} \right) + 5\left( {2x - 1} \right)} \right)$

$

= \left( {2{x^2} + 14x - 6x + 3} \right)\left( {6{x^2} + 42x + 10x - 5} \right) \\

= \left( {2{x^2} + 8x + 3} \right)\left( {6{x^2} + 52 - 5} \right)

$

Now, these two terms $2{x^2} + 8x + 3$ and $6{x^2} + 52x - 5$ are quadratic which can be factored to linear terms.

Now, to factorise these terms find the root using the “Sridharacharya” method then apply this method for $2{x^2} + 8x + 3$.

$

\Rightarrow x = \dfrac{{ - 8 \pm \sqrt {64 - 4 \times 2 \times 3} }}{2} \\

\Rightarrow x = \dfrac{{ - 8 \pm \sqrt {40} }}{2} \\

\Rightarrow x = \dfrac{{2\left( { - 4 \pm \sqrt {10} } \right)}}{2} = - 4 \pm \sqrt {10}

$

So, factor of $2{x^2} + 8x + 3$ is $\left( {x - \left( { - 4 + \sqrt {10} } \right)} \right)$ and $\left( {x - \left( { - 4 - \sqrt {10} } \right)} \right)$.

Now, similarly factorise other terms $6{x^2} + 52x - 5$, by Sridharacharya method.

$

\Rightarrow x = \dfrac{{ - 52 \pm \sqrt {{{\left( {52} \right)}^2} - 4 \times 6 \times \left( { - 5} \right)} }}{2} \\

\Rightarrow x = \dfrac{{ - 52 \pm \sqrt {2704 + 120} }}{2} \\

\Rightarrow x = \dfrac{{ - 52 \pm \sqrt {2824} }}{2} \\

\Rightarrow x = \dfrac{{2\left( { - 26 \pm \sqrt {706} } \right)}}{2} = \left( { - 26 \pm \sqrt {706} } \right)

$

So, factors of $6{x^2} + 52x - 5$ is $\left( {x - \left( { - 26 - \sqrt {706} } \right)} \right)$ and $\left( {x - \left( { - 26 + \sqrt {706} } \right)} \right)$.

Thus, $12{\left( {{x^2} + 7x} \right)^2} - 8\left( {{x^2} + 7x} \right)\left( {2x - 1} \right) - 15{\left( {2x - 1} \right)^2}$ can be factored as

**Therefore $ 12{\left( {{x^2} + 7x} \right)^2} - 8\left( {{x^2} + 7x} \right)\left( {2x - 1} \right) - 15{\left( {2x - 1} \right)^2} = \left( {x - \left( { - 4 - \sqrt {10} } \right)} \right)\left( {x - \left( { - 4 + \sqrt {10} } \right)} \right)\left( {x - \left( { - 26 - \sqrt {706} } \right)} \right)\left( {x - \left( { - 26 + \sqrt {706} } \right)} \right)$**

**Note:**“Sridharacharya” method is an ancient method to find the roots of a quadratic equation. Suppose $a{x^2} + bx + c = 0$ is a quadratic equation and we have to find its roots then, by this method we can write, $x = \dfrac{{ - b \pm \sqrt D }}{2}$ here, $D = {b^2} - 4ac$. And $\left( {x - \left( {\dfrac{{ - b \pm \sqrt D }}{2}} \right)} \right)$ are the factors of the expression $a{x^2} + bx + c$.

Recently Updated Pages

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

Advantages and disadvantages of science

Trending doubts

Which are the Top 10 Largest Countries of the World?

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

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

Give 10 examples for herbs , shrubs , climbers , creepers

Write a letter to the principal requesting him to grant class 10 english CBSE

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

10 examples of evaporation in daily life with explanations