Exercise 8B

Given that \[\mathbf{x}\text{ }\text{ }\mathbf{2}\] and \[\mathbf{x}\text{ }+\text{ }\mathbf{1}\] are factors of f(x) = \[{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{ax}\text{ }+\text{ }\mathbf{b}\]; calculate the values of a and b. Hence, find all the factors of f(x).

Let \[f\left( x \right)\text{ }=\text{ }{{x}^{3}}~+\text{ }3{{x}^{2}}~+\text{ }ax\text{ }+\text{ }b\] As, \[\left( x\text{ }\text{ }2 \right)\] is a factor of f(x), so \[f\left( 2 \right)\text{...

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Using the Remainder Theorem, factorise the expression \[\mathbf{3}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{10}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{x}\text{ }\text{ }\mathbf{6}\]. Hence, solve the equation \[\mathbf{3}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{10}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{x}\text{ }\text{ }\mathbf{6}=0\].

Let’s take f(x) = \[\mathbf{3}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{10}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{x}\text{ }\text{ }\mathbf{6}\] For \[x\text{ }=\text{ }-1\], the value of...

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Using the Factor Theorem, show that: \[\left( \mathbf{3x}\text{ }+\text{ }\mathbf{2} \right)\] is a factor of \[\mathbf{3}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{3x}\text{ }\text{ }\mathbf{2}\]. Hence, factorise the expression \[\mathbf{3}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{3x}\text{ }\text{ }\mathbf{2}\] completely.

Here, f(x) = \[3{{x}^{3}}~+\text{ }2{{x}^{2}}~\text{ }3x\text{ }\text{ }2\] So, \[3x\text{ }+\text{ }2\text{ }=\text{ }0~\Rightarrow x\text{ }=\text{ }-2/3\] Thus, remainder = \[f\left( -2/3...

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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

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{...

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Using the Factor Theorem, show that: \[\left( \mathbf{x}\text{ }\text{ }\mathbf{2} \right)\]is a factor of \[{{\mathbf{x}}^{\mathbf{3}}}~\text{ }\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{9x}\text{ }+\text{ }\mathbf{18}\]. Hence, factorise the expression \[{{\mathbf{x}}^{\mathbf{3}}}~\text{ }\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{9x}\text{ }+\text{ }\mathbf{18}\] completely.

Given, f(x) = \[{{\mathbf{x}}^{\mathbf{3}}}~\text{ }\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{9x}\text{ }+\text{ }\mathbf{18}\] So, \[x\text{ }\text{ }2\text{ }=\text{ }0~\Rightarrow...

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