Question 4 & 5, Review Exercise

Solutions of Question 4 & 5 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.

Is $3 y-2$ a factor of $6 y^{3}-y^{2}-5 y+2$ ?

Solution.

Given $$\begin{align*}3y-2&=0\\ 3y&=2\\ y&=\frac{2}{3}\end{align*}$$ Suppose $$\begin{align*} f(y) &= 6y^{3} - y^{2} - 5y + 2\\ f\left(\frac{2}{3}\right) &= 6\left(\frac{2}{3}\right)^{3} - \left(\frac{2}{3}\right)^{2} - 5\left(\frac{2}{3}\right) + 2 \\ &= 6\left(\frac{8}{27}\right) - \left(\frac{4}{9}\right) - 5\left(\frac{2}{3}\right) + 2 \\ &= \frac{48}{27} - \frac{4}{9} - \frac{10}{3} + 2 \\ &= \frac{16}{9} - \frac{4}{9} - \frac{30}{9} + \frac{18}{9} \\ &= \frac{16 - 4 - 30 + 18}{9} \\ &= \frac{0}{9} = 0. \end{align*}$$

Hence by the factor theorem, $3y - 2$ is a factor of $6y^{3} - y^{2} - 5y + 2$.

If zeros of a polynomial are $4, \frac{3}{5},-2$, find the polynomial.

Solution.

Let the required polynomial be $f(x)$. Given the zeros $4, \frac{3}{5}, -2$, we can write the polynomial as:

$$f(x) = (x - 4)\left(x - \frac{3}{5}\right)(x + 2).$$

Multiplying the factors:

$$\begin{align*} f(x) &= (x - 4) \left( \frac{5x - 3}{5} \right)(x + 2) \\ &= \frac{1}{5}(x - 4)(5x - 3)(x + 2)\\ &=\frac{1}{5} (5x^2 - 3x - 20x + 12 )(x + 2)\\ &=\frac{1}{5} (5x^2 - 23x + 12 )(x + 2)\\ &= \frac{1}{5}\left(5x^3 + 10x^2 - 23x^2 - 46x + 12x + 24\right) \\ &= \frac{1}{5}\left(5x^3 - 13x^2 - 34x + 24\right) \end{align*}$$ Thus, the required polynomial is:

$$f(x) = x^3 - \frac{13}{5}x^2 - \frac{34}{5}x + \frac{24}{5}.$$

< Question 2 & 3 Question 6 & 7>