Question

Read the statements given below:

I. If the polynomial p(x) = 2x³ − kx² + 5x + 2 is exactly divisible by (x + 2), then k = −6.

II. If the polynomial q(x) = x³ − 7x + k when divided by (x − 1) leaves the remainder 2, then k = 6.

III. If two zeros of the polynomial f(x) = x³ − 5x² − 16x + 80 are equal in magnitude and opposite in sigh, then the third zero is 5.

Which of the above statements is not true?

(a) I only
(b) II only
(c) III only
(d) I as well as II

Answer 1

$$\begin{gathered} \left(\mathrm{b}\right)\mathrm{II}\mathrm{only} \\ \left(\text{I}\right)p(-2)=0 \\ =>2\times {(-2)}^3-k{(-2)}^2+5\times (-2)+2=0 \\ =>-16-4k-10+2=0 \\ =>4k=-24 \\ =>k=-6 \\ \therefore \text{I is true}\text{.} \\ \left(\text{II}\right)p\left(1\right)=2 \\ =>1^3-7\times 1-k=2 \\ =>k=-8 \\ \therefore \text{II is false}\text{.} \\ \left(\text{III}\right)\text{Let the zeroes of}f\left(x\right)\text{be}\mathrm{\alpha }\text{,}-\mathrm{\alpha }\mathrm{and}\mathrm{\beta }. \\ \text{Then sum of zeroes=}\frac{-\left(\text{co-efficient of}{x}^2\right)}{\left(\text{co-efficient of}{x}^3\right)}=5 \\ \therefore \mathrm{\alpha }+(-\mathrm{\alpha })+\mathrm{\beta }=5 \\ =>\mathrm{\beta }=5 \\ \begin{array}{l}\text{So the third zero is 5}\text{.}\\ \therefore \text{III is true}\text{.}\end{array} \end{gathered}$$