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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).
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).
Class 10 | Exercise 8B | ICSE | Maths | Remainder and Factor Theorems | Selina
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 \[f\left( 2 \right)\text{ }=\text{ }0\]

\[{{\left( 2 \right)}^{3}}~+\text{ }3{{\left( 2 \right)}^{2}}~+\text{ }a\left( 2 \right)\text{ }+\text{ }b\text{ }=\text{ }0\]

\[8\text{ }+\text{ }12\text{ }+\text{ }2a\text{ }+\text{ }b\text{ }=\text{ }0\]

\[2a+b+20=0\] … (i)

And as, \[\left( x\text{ }+\text{ }1 \right)\] is a factor of f(x), so \[f\left( -1 \right)\text{ }=\text{ }0\]

\[{{\left( -1 \right)}^{3}}~+\text{ }3{{\left( -1 \right)}^{2}}~+\text{ }a\left( -1 \right)\text{ }+\text{ }b\text{ }=\text{ }0\]

\[-1\text{ }+\text{ }3\text{ }\text{ }a\text{ }+\text{ }b\text{ }=\text{ }0\]

\[-a\text{ }+\text{ }b\text{ }+\text{ }2\text{ }=\text{ }0\] … (ii)

Subtracting (ii) from (i), we have

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

3a\text{ }+\text{ }18\text{ }=\text{ }0  \\

a\text{ }=\text{ }-6  \\

\end{array}\]

On substituting the value of a in (ii), we have

\[b\text{ }=\text{ }a\text{ }\text{ }2\text{ }=\text{ }-6\text{ }\text{ }2\text{ }=\text{ }-8\]

Thus, f(x) = \[{{x}^{3}}~+\text{ }3{{x}^{2}}~\text{ }6x\text{ }\text{ }8\]

Now, for \[x\text{ }=\text{ }-1\]

\[f\left( -1 \right)\text{ }=\text{ }{{\left( -1 \right)}^{3}}~+\text{ }3{{\left( -1 \right)}^{2}}~\text{ }6\left( -1 \right)\text{ }\text{ }8\text{ }=\text{ }-1\text{ }+\text{ }3\text{ }+\text{ }6\text{ }\text{ }8\text{ }=\text{ }0\]

Therefore, \[\left( x\text{ }+\text{ }1 \right)\] is a factor of f(x).

Now, performing long division we have

Hence,

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

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

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

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

i

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