Factorise the following
$12{\left( {{x^2} + 7x} \right)^2} - 8\left( {{x^2} + 7x} \right)\left( {2x - 1} \right) - 15{\left( {2x - 1} \right)^2}$.
Answer
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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 $a{x^2} + bx + c$.
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 $a{x^2} + bx + c$.
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