Question 6 & 7, Review Exercise

Solutions of Question 6 & 7 of Review Exercise of Unit 05: Polynomials. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Find the value of ' $k$ ' so that the remainder upon dividing $\left(x^{2}+8 x+k\right)$ by $(x-4)$ is zero.

Solution.

Let \( p(x) = x^{2} + 8x + k \). We are given that the remainder upon dividing \( p(x) \) by \( (x - 4) \) is zero.
By the remainder theorem, the remainder is \( p(4) \).

Since \( p(4) = 0 \), we have: \begin{align*} p(4) &= (4)^2 + 8(4) + k \\ &= 16 + 32 + k \\ &= 48 + k. \end{align*}

For the remainder to be zero:

\[ 48 + k = 0. \]

Thus,

\[ k = -48. \]

Suppose that the quotient upon dividing one polynomial by another is $3 x^{2}-x+32-\frac{121}{x+4}$. What is the dividend?

Solution.

The question doesn't seem to solvable.

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