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ICSE Class 10 Selina Maths Remainder and Factor Theorems
Remainder and Factor Theorems

Find the value of ‘m’, if \[\mathbf{m}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{3}\] and \[{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{mx}\text{ }+\text{ }\mathbf{4}\] leave the same remainder when each is divided by \[\mathbf{x}\text{ }\text{ }\mathbf{2}\].

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Let f(x) = \[\mathbf{m}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{3}\] and g(x) = \[{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{mx}\text{ }+\text{...

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If \[\mathbf{x}\text{ }\text{ }\mathbf{2}\] is a factor of \[{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{ax}\text{ }+\text{ }\mathbf{b}\] and \[\mathbf{a}\text{ }+\text{ }\mathbf{b}\text{ }=\text{ }\mathbf{1}\], find the values of a and b.

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Let f(x) = \[{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{ax}\text{ }+\text{ }\mathbf{b}\] Given,  \[\left( x\text{ }\text{ }2 \right)\]is a factor of f(x). Then, remainder = \[f\left( 2...

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If \[\left( \mathbf{x}\text{ }+\text{ }\mathbf{1} \right)\] and \[\left( \mathbf{x}\text{ }\text{ }\mathbf{2} \right)\]are factors of \[{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\left( \mathbf{a}\text{ }+\text{ }\mathbf{1} \right){{\mathbf{x}}^{\mathbf{2}}}~\text{ }\left( \mathbf{b}\text{ }\text{ }\mathbf{2} \right)\mathbf{x}\text{ }\text{ }\mathbf{6}\], find the values of a and b. And then, factorise the given expression completely.

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Let’s take f(x) = \[{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\left( \mathbf{a}\text{ }+\text{ }\mathbf{1} \right){{\mathbf{x}}^{\mathbf{2}}}~\text{ }\left( \mathbf{b}\text{ }\text{ }\mathbf{2}...

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What should be subtracted from \[\mathbf{3}{{\mathbf{x}}^{\mathbf{3}}}~\text{ }\mathbf{8}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{4x}\text{ }\text{ }\mathbf{3}\], so that the resulting expression has \[\mathbf{x}\text{ }+\text{ }\mathbf{2}\] as a factor?

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Let’s assume the required number to be k. And let f(x) = \[3{{x}^{3}}~\text{ }8{{x}^{2}}~+\text{ }4x\text{ }\text{ }3\text{ }\text{ }k\] From the question, we have \[f\left( -2 \right)\text{...

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When \[{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{mx}\text{ }+\text{ }\mathbf{4}\]is divided by \[\mathbf{x}\text{ }\text{ }\mathbf{2}\], the remainder is \[\mathbf{m}\text{ }+\text{ }\mathbf{3}\]. Find the value of m.

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Let us consider f(x) = \[{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{mx}\text{ }+\text{ }\mathbf{4}\] From the question, we have \[f\left( 2...

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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.

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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...

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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).

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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 \[\mathbf{3}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{10}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{x}\text{ }\text{ }\mathbf{6}=0\].

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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.

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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

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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.

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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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Find the value of k, if \[\mathbf{2x}\text{ }+\text{ }\mathbf{1}\] is a factor of \[\left( \mathbf{3k}\text{ }+\text{ }\mathbf{2} \right){{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\left( \mathbf{k}\text{ }\text{ }\mathbf{1} \right)\].

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Let us consider f(x) = \[\left( \mathbf{3k}\text{ }+\text{ }\mathbf{2} \right){{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\left( \mathbf{k}\text{ }\text{ }\mathbf{1} \right)\] Now, \[2x\text{ }+\text{...

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Use the Remainder Theorem to find which of the following is a factor of \[\mathbf{2}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{5x}\text{ }\text{ }\mathbf{6}\]. (i) \[\mathbf{x}\text{ }+\text{ }\mathbf{1}\] (ii) \[\mathbf{2x}\text{ }\text{ }\mathbf{1}\] (iii) \[\mathbf{x}\text{ }+\text{ }\mathbf{2}\]

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According to  remainder theorem, when a polynomial f (x) is divided by x – a, then the remainder is f(a). Here, f(x) = \[\mathbf{2}{{\mathbf{x}}^{\mathbf{3}}}~+\text{...

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