Question

Match of the following columns:

Column I		Column II
(a)	When p(x) = 81x4 + 54x3 − 9x2 − 3x + 2 is divided by (3x + 2), then reaminder ....... .	(p)	1
(b)	p(x) = ax3 + 3x2 − 3 and q(x) = 2x3 − 5x + a divided by (x − 4) leave the same remainder. Then, a ...... .	(q)	−7
(c)	If $p\left(x\right)=7x^2-4\sqrt{2}x+c$is completely divisible by $\left(x-\sqrt{2}\right)$, then c = ...... .	(r)	0
(d)	If q(x) = 2x3 + bx2 + 11x + b + 3 is divisibl by (2x + 1), then b = ....... .	(s)	−6

(a) ....... .
(b) ....... .
(c) ....... .
(d) ....... .

Answer 1

(a)
Let:
$p\left(x\right)=81x^4+54x^3-9x^2-3x+2$
$3x+2=0\Rightarrow x=\frac{-2}{3}$
By the remainder theorem, we know that when $p\left(x\right)\text{is divided by}\left(3x+2\right),\text{the remainder is}p\left(\frac{-2}{3}\right).$
Thus, we have:
$p\left(\frac{-2}{3}\right)=81\times {\left(\frac{-2}{3}\right)}^4+54\times {\left(\frac{-2}{3}\right)}^3-9\times {\left(\frac{-2}{3}\right)}^2-3\times \left(\frac{-2}{3}\right)+2$
$$\begin{gathered} =\left(81\times \frac{16}{81}\right)-\left(54\times \frac{8}{27}\right)-\left(9\times \frac{4}{9}\right)+2+2 \\ =\left(16-16-4+2+2\right) \\ =0 \end{gathered}$$
∴ $\left(a\right)\Rightarrow \left(r\right)$

(b)
Let:
$p\left(x\right)=a{x}^3+3x^2-3$
$x-4=0\Rightarrow x=4$
By the remainder theorem, we know that when $p\left(x\right)\text{is divided by}\left(x-4\right),\text{the remainder is}p\left(4\right).$
Thus, we have:
$$\begin{gathered} p\left(4\right)=a\times 4^3+3\times 4^2-3 \\ =64a+48-3 \\ =64a+45 \end{gathered}$$

Also,
Let:
$q\left(x\right)=2x^3-5x+a$
By the remainder theorem, we know that when​ $q\left(x\right)\text{is divided by}\left(x-4\right),\text{the remainder is}q\left(4\right).$
Thus, we have:
$$\begin{gathered} q\left(4\right)=2\times 4^3-5\times 4+a \\ =128-20+a \\ =108+a \end{gathered}$$
Now,
$$\begin{gathered} 64a+45=108+a \\ \Rightarrow 63a=63 \\ \Rightarrow a=1 \end{gathered}$$
∴ $\left(b\right)\Rightarrow \left(p\right)$

(c)
Let:
$p\left(x\right)=7x^2+-4\sqrt{2}x+c$
Now,
$x-\sqrt{2}=0\Rightarrow x=\sqrt{2}$
$$\begin{gathered} p\left(x\right)\text{is completely divisible by}\left(x-\sqrt{2}\right). \\ \Rightarrow p\left(\sqrt{2}\right)=0 \\ \Rightarrow 7\times {\left(\sqrt{2}\right)}^2+\left(-4\sqrt{2}\right)\times \sqrt{2}+c=0 \\ \Rightarrow 14-8+c=0 \\ \Rightarrow 6+c=0 \\ \Rightarrow c=-6 \end{gathered}$$
∴ $\left(c\right)\Rightarrow \left(s\right)$

(d)
Let:
$q\left(x\right)=2x^3+b{x}^2+11x+b+3$
Now,
$2x-1=0\Rightarrow x=\frac{1}{2}$
$$\begin{gathered} q\left(x\right)\text{is completely divisible by}\left(2x-1\right). \\ \Rightarrow q\left(\frac{1}{2}\right)=0 \\ \Rightarrow 2\times {\left(\frac{1}{2}\right)}^3+b\times {\left(\frac{1}{2}\right)}^2+11\times \left(\frac{1}{2}\right)+b+3=0 \\ \Rightarrow \frac{1}{4}+\frac{b}{4}+\frac{11}{2}+b+3=0 \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{1+b+22+4b+12}{4}=0 \\ \Rightarrow 35+5b=0 \\ \Rightarrow b=\frac{-35}{5} \\ \Rightarrow b=-7 \end{gathered}$$
∴ $\left(d\right)\Rightarrow \left(q\right)$