Question

When $ 3{x^2} + 6x - 10 $ is divided by $ x + k $ , the remainder is 14. How do you determine the value of k?

Answer 1

Hint: In this question, we are given the dividend, divisor and the remainder and we have to find the value of k so we will use the division algorithm. But we also don’t know the value of the quotient so we will have so three unknown quantities and only one equation, so to find the value of k, we will find the condition where the value of divisor becomes zero.

Complete step-by-step answer:
We know that $ a = bq + r $
We have $ a = 3{x^2} + 6x - 10 $ and $ b = x + k $
When $ b = 0,\,a = r $
So, at
 $
  x = - k,\,a = 14 \\
   \Rightarrow 3{( - k)^2} + 6( - k) - 10 = 14 \\
   \Rightarrow 3{k^2} - 6k - 24 = 0 \;
  $
The obtained equation is a quadratic equation as the degree of the equation is 2, so it can be solved by factorization as follows –
 $
  3({k^2} - 2k - 8) = 0 \\
   \Rightarrow {k^2} - 4k + 2k - 8 = 0 \\
   \Rightarrow k(k - 4) + 2(k - 4) = 0 \\
   \Rightarrow (k - 4)(k + 2) = 0 \\
   \Rightarrow k = 4,\,k = - 2 \;
  $
Hence when $ 3{x^2} + 6x - 10 $ is divided by $ x + k $ , and the remainder is 14, we get $ k = 4 $ or $ k = - 2 $ .
So, the correct answer is “ $ k = 4 $ or $ k = - 2 $ ”.

Note: We are given a fraction whose numerator and denominator are polynomials. When a polynomial equation of degree m divides another polynomial of degree n such that $ m \leqslant n $ , then the degree of the quotient is $ n - m $ and the degree of the remainder is $ m - 1 $ . So the quotient of the above expression will have degree 1 and the remainder is a constant. We have obtained a polynomial equation in the end, in which the highest power of x is 2, so the degree of this equation is 2 and thus it has 2 solutions. We solved the equation by using factorization, or we can also use the quadratic formula if factorization is not possible.