\CIE\0606\2020\w\paper 12 \no 10
The polynomial $\mathrm{p}(x)=6 x^{3}+a x^{2}+b x+2$, where $a$ and $b$ are integers, has a factor of $x-2$.
(a) Given that $\mathrm{p}(1)=-2 \mathrm{p}(0)$, find the value of $a$ and of $b$.
(b) Using your values of $a$ and $b$,
(i) find the remainder when $\mathrm{p}(x)$ is divided by $2 x-1$,
(ii) factorise $\mathrm{p}(x)$.
|
*********math solution************
********************************************** a) $\begin{array}[t]{ll}&p(1)=6(1)^{3}+a(1)^{2}+b(1)+2=8+a+b \\&p(0)=6(0)^{3}+a(0)^{2}+b(0)+2=2 \\&p(2)=6(2)^{3}+a(2)^{2}+b(2)+2=48+4 a+2 b+2=50+4 a+2 b \\&p(1)=-2 p(0) \\&8+a+b=-2(2)=-4 \\&a+b=-12 \quad \ldots \quad(1) \\&p(2)=0=50+4 a+2 b \\&2 a+b=-25 \quad \ldots \text { (2) } \\&(2)-(1): a=-13 \\&b=1\end{array}$ b)$\begin{array}[t]{ll}&\text { Hence } P(x)=6 x^{3}-13 x^{2}+x+2 \\&\text { i) } \quad \text { Remainder when } p(x) \text { is divided by } 2 x-1 \text { , } \\&=p\left(\frac{1}{2}\right) \\&=6\left(\frac{1}{2}\right)^{3}-13\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)+2 \\&=0 \end{array}$ $\begin{array}[t]{ll}\text{ii) }p(x)=(x-2)(2 x-1)(c x+d)=6 x^{3}-13 x^{2}+x+2 \\c=3, d=1 \\\therefore p(x)=(x-2)(2 x-1)(3 x+1)\end{array}$ ********************************************* **********end math solution********************$ |
|---|
Post a Comment