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{...
Factorise the expression $f(x)=2 x^{3}-7 x^{2}-3 x+18$. Hence, find all possible values of $x$ for which $f(x)=$ 0
Given: f(x) = 2x3 – 7x2 – 3x + 18 By hit and trial method For x = 2, the value of f(x) will be f(2) = 2(2)3 – 7(2)2 – 3(2) + 18 = 16 – 28 – 6 + 18 = 0 As f(2) = 0, (x – 2) is a factor of f(x). Now,...
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...
Using the Remainder Theorem, factorise each of the following completely. \[{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{4x}\text{ }\text{ }\mathbf{4}\]
Let f(x) = \[{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{4x}\text{ }\text{ }\mathbf{4}\] For \[x\text{ }=\text{ }-1\], the value of f(x) will be...
Using the Remainder Theorem, factorise each of the following completely. \[\mathbf{4}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{7}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{36x}\text{ }\text{ }\mathbf{63}\]
Let f(x) = \[\mathbf{4}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{7}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{36x}\text{ }\text{ }\mathbf{63}\] For \[x\text{ }=\text{ }3\], the value of f(x)...
Using the Remainder Theorem, factorise each of the following completely. $3 x^{3}+2 x^{2}-23 x-30$
Given: f(x) = 2x3 + x2 – 13x + 6 By hit and trial we put x=-2 For x = -2, the value of f(x) will be f(-2) = 3(-2)3 + 2(-2)2 – 23(-2) – 30 = -24 + 8 + 46 – 30 = -54 + 54 = 0 As f(-2) = 0, so (x + 2)...
Using the Remainder Theorem, factorise each of the following completely. \[\mathbf{2}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{13x}\text{ }+\text{ }\mathbf{6}\]
Let f(x) = \[\mathbf{2}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{13x}\text{ }+\text{ }\mathbf{6}\] For \[x\text{ }=\text{ }2\], the value of f(x) will...
Using the Remainder Theorem, factorise each of the following completely. \[\mathbf{3}{{\mathbf{x}}^{\mathbf{3}~}}+\text{ }\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}~-\text{ }\mathbf{19x}\text{ }+\text{ }\mathbf{6}\]
Let f(x) = \[\mathbf{3}{{\mathbf{x}}^{\mathbf{3}~}}+\text{ }\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}~-\text{ }\mathbf{19x}\text{ }+\text{ }\mathbf{6}\] For x=2, the value of f(x) will be...
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...
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{...
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...