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Using the Factor Theorem, show that: \[\left( \mathbf{x}\text{ }+\text{ }\mathbf{5} \right)\] is a factor of \[\mathbf{2}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{5}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{28x}\text{ }\text{ }\mathbf{15}\]. Hence, factorise the expression \[\mathbf{2}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{5}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{28x}\text{ }\text{ }\mathbf{15}\] completely
Using the Factor Theorem, show that: \[\left( \mathbf{x}\text{ }+\text{ }\mathbf{5} \right)\] is a factor of \[\mathbf{2}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{5}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{28x}\text{ }\text{ }\mathbf{15}\]. Hence, factorise the expression \[\mathbf{2}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{5}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{28x}\text{ }\text{ }\mathbf{15}\] completely
Class 10 | Exercise 8B | ICSE | Maths | Remainder and Factor Theorems | Selina
Using the Factor Theorem, show that: \[\left( \mathbf{x}\text{ }+\text{ }\mathbf{5} \right)\] is a factor of \[\mathbf{2}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{5}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{28x}\text{ }\text{ }\mathbf{15}\]. Hence, factorise the expression \[\mathbf{2}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{5}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{28x}\text{ }\text{ }\mathbf{15}\] completely

Given, f(x) = \[\mathbf{2}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{5}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{28x}\text{ }\text{ }\mathbf{15}\]

So, \[x\text{ }+\text{ }5\text{ }=\text{ }0~\Rightarrow x\text{ }=\text{ }-5\]

Thus, remainder = \[f\left( -5 \right)\]

\[\begin{array}{*{35}{l}}

=\text{ }2{{\left( -5 \right)}^{3}}~+\text{ }5{{\left( -5 \right)}^{2}}~\text{ }28\left( -5 \right)\text{ }\text{ }15  \\

=\text{ }-250\text{ }+\text{ }125\text{ }+\text{ }140\text{ }\text{ }15  \\

=\text{ }-265\text{ }+\text{ }265  \\

=\text{ }0  \\

\end{array}\]

Therefore, \[\left( x\text{ }+\text{ }5 \right)\] is a factor of f(x).

Now, performing division of polynomial f(x) by \[\left( x\text{ }+\text{ }5 \right)\]we get

So, \[2{{x}^{3}}~+\text{ }5{{x}^{2}}~\text{ }28x\text{ }\text{ }15\text{ }=\text{ }\left( x\text{ }+\text{ }5 \right)\text{ }(2{{x}^{2}}~\text{ }5x\text{ }\text{ }3)\]

Further, on factorisation

\[\begin{array}{*{35}{l}}

=\text{ }\left( x\text{ }+\text{ }5 \right)\text{ }[2{{x}^{2}}~\text{ }6x\text{ }+\text{ }x\text{ }\text{ }3]  \\

=\text{ }\left( x\text{ }+\text{ }5 \right)\text{ }\left[ 2x\left( x\text{ }\text{ }3 \right)\text{ }+\text{ }1\left( x\text{ }\text{ }3 \right) \right]\text{ }=\text{ }\left( x\text{ }+\text{ }5 \right)\text{ }\left( 2x\text{ }+\text{ }1 \right)\text{ }\left( x\text{ }\text{ }3 \right)  \\

\end{array}\]

Thus, f(x) is factorised as \[\left( x\text{ }+\text{ }5 \right)\text{ }\left( 2x\text{ }+\text{ }1 \right)\text{ }\left( x\text{ }\text{ }3 \right)\]

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