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Using the Remainder Theorem, factorise each of the following completely.

    \[\mathbf{2}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{13x}\text{ }+\text{ }\mathbf{6}\]

Class 10 | Exercise 8B | ICSE | Maths | Remainder and Factor Theorems | Selina
Using the Remainder Theorem, factorise each of the following completely.

    \[\mathbf{2}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{13x}\text{ }+\text{ }\mathbf{6}\]

 

Let f(x) =

    \[\mathbf{2}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{13x}\text{ }+\text{ }\mathbf{6}\]

For

    \[x\text{ }=\text{ }2\]

, the value of f(x) will be

    \[f\left( 2 \right)\text{ }=\text{ }2{{\left( 2 \right)}^{3}}~+\text{ }{{\left( 2 \right)}^{2}}~\text{ }13\left( 2 \right)\text{ }+\text{ }6\text{ }=\text{ }16\text{ }+\text{ }4\text{ }\text{ }26\text{ }+\text{ }6\text{ }=\text{ }0\]

As

    \[f\left( 2 \right)\text{ }=\text{ }0\]

, so

    \[\left( x\text{ }\text{ }2 \right)\]

is a factor of f(x).

Now, performing long division we have

Therefore,

    \[\begin{array}{*{35}{l}} ,\text{ }f\left( x \right)\text{ }=\text{ }\left( x\text{ }-2 \right)\text{ }(2{{x}^{2}}~+\text{ }5x\text{ }\text{ }3) \\ =\text{ }\left( x\text{ }\text{ }2 \right)\text{ }[2{{x}^{2}}~+\text{ }6x\text{ }\text{ }x\text{ }\text{ }3] \\ =\text{ }\left( x\text{ }\text{ }2 \right)\text{ }\left[ 2x\left( x\text{ }+\text{ }3 \right)\text{ }-1\left( x\text{ }+\text{ }3 \right) \right] \\ =\text{ }\left( x\text{ }\text{ }2 \right)\text{ }\left[ 2x\left( x\text{ }+\text{ }3 \right)\text{ }-1\left( x\text{ }+\text{ }3 \right) \right] \\ =\text{ }\left( x\text{ }\text{ }2 \right)\text{ }\left( 2x\text{ }\text{ }1 \right)\text{ }\left( x\text{ }+\text{ }3 \right) \\ \end{array}\]

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