Question

The remainder when $x^5+K{x}^2$ is divided by $(x-1)(x-2)(x-3)$ contains no term in $x^2$. Then the value of $K$ is

• A. $-90$
• B. $90$
• C. $74$
• D. $15$

Answer 1

The correct option is A $-90$

$(x–1)(x–2)(x–3)=x^3–6x^2+11x–6$

Let the quotient and remainder when

$x^5+k{x}^2$ is divided by $(x–1)(x–2)(x–3)$ be

$(a{x}^2+bx+c)$ and $dx+e$

Now

$x^5+k{x}^2=(x^3–6x^2+11x–6)(a{x}^2+bx+c)+dx+e$

Comparing the coefficients of $x^5,x^4,x^{3}and{x}^2$ we get

$a=1$

b – 6 a = 0 $\Rightarrow$ b = 6

c – 6 b + 11 a = 0 $\Rightarrow$ c = 25

$–6c+11b–6a=k$

$\Rightarrow$ k = –90

Hence, the value of k is –90