Question

If the polynomial f(x) = ${x^4} - 6{x^3} + 16{x^2} - 25x + 10$ is divided by another polynomial, ${x^2} - 2x + k$, the remainder comes out to be (x + a), then values of k and a are,
$\left( a \right)k = - 2{\text{ and }}a = 4$
$\left( b \right)k = 5{\text{ and }}a = - 5$
$\left( c \right)k = - 3{\text{ and }}a = - 7$
$\left( d \right)$ None of these.

Answer 1

Hint: In this particular question use the concept that if we divide fourth order polynomial by second order polynomial the quotient will be a second order polynomial so assume any general second order polynomial be the quotient (say, $p{x^2} + qx + r$) where p, q, and r are the constant real parameters, so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given polynomial
f(x) = ${x^4} - 6{x^3} + 16{x^2} - 25x + 10$
Now when this polynomial is divided by another polynomial ${x^2} - 2x + k$, the remainder comes out to be (x + a).
Let $\dfrac{{f\left( x \right)}}{{{x^2} - 2x + k}} = Q\dfrac{R}{D}$.............. (1)
Where, Q = quotient, R = remainder and D = divisor.
As the highest power of the given polynomial is four, so it is a fourth order polynomial.
So when this polynomial is divided by a quadratic polynomial then the quotient must be a quadratic polynomial.
So let, Q = $p{x^2} + qx + r$, where p, q and r, are constant real parameters and it is given that R = (x + a) and D = ${x^2} - 2x + k$
Now from equation (1) we have,
$ \Rightarrow \dfrac{{{x^4} - 6{x^3} + 16{x^2} - 25x + 10}}{{{x^2} - 2x + k}} = \left( {p{x^2} + qx + r} \right)\dfrac{{\left( {x + a} \right)}}{{\left( {{x^2} - 2x + k} \right)}}$
Now simplify we have,
\[ \Rightarrow {x^4} - 6{x^3} + 16{x^2} - 25x + 10 = \left( {p{x^2} + qx + r} \right)\left( {{x^2} - 2x + k} \right) + \left( {x + a} \right)\]
\[ \Rightarrow {x^4} - 6{x^3} + 16{x^2} - 25x + 10 = \left( {p{x^4} - \left( {2p - q} \right){x^3} + \left( {pk - 2q + r} \right){x^2} + \left( {qk - 2r + 1} \right)x + rk + a} \right)\]
Now on comparing we have,
\[ \Rightarrow p = 1\]
\[ \Rightarrow 2p - q = 6\]................ (1)
\[ \Rightarrow pk - 2q + r = 16\]................ (2)
\[ \Rightarrow qk - 2r + 1 = - 25\]............. (3)
\[ \Rightarrow rk + a = 10\]............... (4)
Now substitute the value of p in equation (1) we have,
\[ \Rightarrow 2\left( 1 \right) - q = 6\]
\[ \Rightarrow q = - 6 + 2 = - 4\]
Now substitute the value of p and q in equation (2) we have,
\[ \Rightarrow k - 2\left( { - 4} \right) + r = 16\]
\[ \Rightarrow k + r = 16 - 8 = 8\].................. (5)
Now substitute the value of q in equation (3) we have,
\[ \Rightarrow - 4k - 2r + 1 = - 25\]
\[ \Rightarrow - 4k - 2r = - 26\]
Divide by -2 throughout we have,
\[ \Rightarrow 2k + r = 13\].................. (6)
Now subtract equation (5) and (6) we have,
$ \Rightarrow k + r - 2k - r = 8 - 13$
$ \Rightarrow k = 5$
Now from equation (5) we have,
\[ \Rightarrow 5 + r = 8\]
\[ \Rightarrow r = 8 - 5 = 3\]
Now substitute the value of r and k in equation (4) we have,
\[ \Rightarrow \left( 3 \right)\left( 5 \right) + a = 10\]
\[ \Rightarrow a = 10 - 15\]
\[ \Rightarrow a = - 5\]
So this is the required value of a and k.

So, the correct answer is “Option b”.

Note: Whenever we face such types of questions the key concept we have to remember is that when we multiply the divisor polynomial by the quotient polynomial and add the remainder in them we will get the resultant polynomial, so on comparing we get some equation with unknown parameters, simply solve these equation we will get the required answer.