Question

Find cubic polynomial whose zeroes are 3, $\frac{1}{2}$ & $-1$

Answer 1

Given: zeroes of a cubic polynomial $3,\frac{1}{2},-1$

By factor theorem,

$(x-3)(x-\frac{1}{2})(x+1)$ are the factors of the cubic polynomial.

We know that $(x+a)(x+b)=x^2+(a+b)x+ab$

Using the factors $(x-3)(x-\frac{1}{2})$ we have as

$x^2+(-3-\frac{1}{2})x+\frac{3}{2}$

$=x^2-\frac{-7}{2}x+\frac{3}{2}$

Again $(x+1)(x^2-\frac{7}{2}x+\frac{3}{2})=x^3-\frac{7x^2}{2}+\frac{3x}{2}+x^2-\frac{7}{2}x+\frac{3}{2}$

$=x^3+\frac{2x^2-7x^2}{2}+\frac{3x-7x}{2}+\frac{3}{2}$

$=x^3-\frac{5x^2}{2}-2x+\frac{3}{2}$

$=2x^3-5x^2-4x+3$ is the required cubic polynomial.