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Show that

    \[\left( \mathbf{x}\text{ }\text{ }\mathbf{1} \right)\]

is a factor of

    \[{{\mathbf{x}}^{\mathbf{3}}}~\text{ }\mathbf{7}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{14x}\text{ }\text{ }\mathbf{8}\]

. Hence, completely factorise the given expression.
Class 10 | Exercise 8C | ICSE | Maths | Remainder and Factor Theorems | Selina
Show that

    \[\left( \mathbf{x}\text{ }\text{ }\mathbf{1} \right)\]

is a factor of

    \[{{\mathbf{x}}^{\mathbf{3}}}~\text{ }\mathbf{7}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{14x}\text{ }\text{ }\mathbf{8}\]

. Hence, completely factorise the given expression.

Let us consider f(x) =

    \[{{\mathbf{x}}^{\mathbf{3}}}~\text{ }\mathbf{7}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{14x}\text{ }\text{ }\mathbf{8}\]

Then, for

    \[x\text{ }=\text{ }1\]

    \[f\left( 1 \right)\text{ }=\text{ }{{\left( 1 \right)}^{3}}~\text{ }7{{\left( 1 \right)}^{2}}~+\text{ }14\left( 1 \right)\text{ }\text{ }8\text{ }=\text{ }1\text{ }\text{ }7\text{ }+\text{ }14\text{ }\text{ }8\text{ }=\text{ }0\]

Thus,

    \[\left( \mathbf{x}\text{ }\text{ }\mathbf{1} \right)\]

is a factor of f(x).

Now, performing long division we have

Hence,

    \[f\left( x \right)\text{ }=\text{ }\left( x\text{ }\text{ }1 \right)\text{ }({{x}^{2}}~\text{ }6x\text{ }+\text{ }8)\]

    \[=\text{ }\left( x\text{ }\text{ }1 \right)\text{ }({{x}^{2}}~\text{ }4x\text{ }\text{ }2x\text{ }+\text{ }8)\]

    \[=\text{ }\left( x\text{ }\text{ }1 \right)\text{ }\left[ x\left( x\text{ }\text{ }4 \right)\text{ }-2\left( x\text{ }\text{ }4 \right) \right]\]

    \[=\text{ }\left( x\text{ }\text{ }1 \right)\text{ }\left( x\text{ }\text{ }4 \right)\text{ }\left( x\text{ }\text{ }2 \right)\]

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