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{...
Factorise \[{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{6}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{11x}\text{ }+\text{ }\mathbf{6}\]completely using factor theorem.
Let f(x) = \[{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{6}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{11x}\text{ }+\text{ }\mathbf{6}\] For \[x\text{ }=\text{ }-1\], the value of f(x) is...
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.
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...
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.
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}...
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?
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{...
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.
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...
Using Remainder Theorem, factorise: \[{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{10}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{37x}\text{ }+\text{ }\mathbf{26}\]completely.
Let f(x) = \[{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{10}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{37x}\text{ }+\text{ }\mathbf{26}\] According to remainder theorem, we know that For \[x\text{...
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...