Question

Find the zeroes of the polynomial $f\left(x\right)=x^3-8x^2+19x-12,\text{if it is given}\text{that the product of its two zereoes is 4.}$

• A.
  3,4 and 2
• B.
  3,4 and 1
• C.
  3,2 and 1
• D.
  3,1 and 7

Answer 1

The correct option is B

3,4 and 1
$\text{Let zeroes of}f\left(x\right)=x^3-8x^2+19x-12be\alpha .\beta and\gamma .\alpha \beta =4\alpha \beta \gamma =\frac{-d}{a}=\frac{-(-12)}{1}=12\Rightarrow 4\gamma =12\Rightarrow \gamma =\frac{12}{4}=3\therefore (x-3)\text{is a factor of}f\left(x\right)$

$x^3-8x^2+19x-12\Rightarrow (x-3)(x^2-5x+4)\Rightarrow (x-3)(x^2-4x-x+4)\Rightarrow (x-3)\left[x\right(x-4)-1(x-4\left)\right]\Rightarrow (x-3)(x-4)(x-1)\text{Zeroes of f(x) are 3, 4, and 1}$
So, the correct option is b