Question

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Q. If (x - 2) and (x - 3) are factors of a polynomial $x^3+a{x}^2+bx+6$, find a and b.

Answer 1

Dear student
$$\begin{gathered} \mathrm{Given}:\left(\mathrm{x}-2\right)\mathrm{and}\left(\mathrm{x}-3\right)\mathrm{are}\mathrm{factors}\mathrm{of}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3+{\mathrm{ax}}^2+\mathrm{bx}+6 \\ \Rightarrow \mathrm{f}\left(2\right)=0=\mathrm{f}\left(3\right) \\ \mathrm{f}\left(2\right)=2^3+\mathrm{a}{\left(2\right)}^2+2\mathrm{b}+6 \\ =8+4\mathrm{a}+2\mathrm{b}+6 \\ =4\mathrm{a}+2\mathrm{b}+14 \\ \mathrm{So},4\mathrm{a}+2\mathrm{b}+14=0 \\ \Rightarrow 4\mathrm{a}+2\mathrm{b}=-14 \\ \Rightarrow 2a+b=-7 \\ \Rightarrow b=-7-2a...\left(1\right) \\ \mathrm{and}\mathrm{f}\left(3\right)=3^3+\mathrm{a}{\left(3\right)}^2+3\mathrm{b}+6 \\ =27+9\mathrm{a}+3\mathrm{b}+6 \\ =9\mathrm{a}+3\mathrm{b}+33 \\ \mathrm{So},9\mathrm{a}+3\mathrm{b}+33=0 \\ \Rightarrow 9\mathrm{a}+3\mathrm{b}=-33 \\ \Rightarrow 3a+b=-11 \\ \Rightarrow 3a-7-2a=-11[u\sin g1] \\ \Rightarrow a=-4 \\ andb=-7-2(-4)=-7+8=1 \\ So,\boxed{\mathbf{a}\mathbf{=}\mathbf{-}\mathbf{4}\mathbf{}\mathbf{a}\mathbf{n}\mathbf{d}\mathbf{}\mathbf{b}\mathbf{=}\mathbf{1}} \end{gathered}$$
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