Question

The cubic polynomial $p\left(x\right)$ satisfies $p\left(0\right)=1,p\left(1\right)=20,p\left(2\right)=40,p\left(3\right)=60.$ Then the value of $p\left(4\right)$ is

• A. $78$
• B. $81$
• C. $79$
• D. $80$

Answer 1

The correct option is C $79$

Given cubic polynomial
$p\left(x\right)=a{x}^3+b{x}^2+cx+d$
Also,
$p\left(0\right)=1\Rightarrow d=1$

$p\left(1\right)=20$
$\Rightarrow a+b+c+d=20\Rightarrow a+b+c=19$
$\Rightarrow c=19-a-b$ $.....\left(1\right)$

$p\left(2\right)=40$
$\Rightarrow 8a+4b+2c+1=40\Rightarrow 8a+4b+2(19-a-b)=39$
$\Rightarrow 6a+2b=1\Rightarrow 3a+b=\frac{1}{2}$ $....\left(2\right)$

$p\left(3\right)=60$
$\Rightarrow 27a+9b+3c+1=60$
$\Rightarrow 9(3a+b)+3c=59$
$\Rightarrow \frac{9}{2}+3c=59\Rightarrow c=\frac{109}{6}$
putting the value of $c=\frac{109}{6}$ in eq.(1) we get,
$a+b=19-c\Rightarrow a+b=19-\frac{109}{6}$
$\Rightarrow a+b=\frac{5}{6}$
$\Rightarrow 3a+3b=\frac{5}{2}$ $.....\left(3\right)$
substracting eq.$\left(2\right)$ from $\left(3\right)$ we get,
$2b=2\Rightarrow b=1$

Re-substituting value of $b=1$ in equation$\left(2\right)$ we get,
$3a=-\frac{1}{2}\Rightarrow a=-\frac{1}{6}$
substituting value of $a,b,c,$ and $d$ in the cubic polynomial we get
$p\left(4\right)=64a+16b+4c+d=64(-\frac{1}{6})+16+4\left(\frac{109}{6}\right)+1=79$