Question

find the zeros of the polynomial f(x)=x³-13x²+52x-60,if it is given that the product of its 2 zeroes is 10.

Answer 1

$let \: the \: zeros \: of \: polynomial \: \\ are \: \alpha \: \: \: \beta \: \: \: \gamma \\ \alpha \beta \gamma = \frac{ - d}{a} = 60 \\ 10 \gamma = 60 \\ \gamma = 6 \: \: \: \: ans \\ \alpha + \beta + \gamma = \frac{ - b}{a} = 13 \\ \alpha + \beta = 13 - 6 = 7 \: \: \: \: \: eq1 \\ \alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a} = 52 \\ 10 + 6 \beta + 6 \alpha = 52 \\ 6 \alpha + 6 \beta = 42 \\ \alpha + \beta = 7 \: \: \: \: eq2 \\ from \: eq 1\: and \: eq2 \: we \: cannot \\ \: find \: the \: value \: of \: \alpha \: and \: \beta .$
$from \: one \: zero \: make \: factor \: \\ x - 6 \: \: and \: divide \: f(x) \: by \: that \: factor \\ \\ \frac{ {x}^{3} - 13 {x}^{2} + 52x - 60 }{x - 6} \\ = {x}^{2} - 7x + 10 \\ now \: factorise \: this \\ = {x}^{2} - 5x - 2x + 10 \\ = x(x - 5) - 2(x - 5) \\ = (x - 5)(x - 2) \\ x = 5 \\ x = 2 \\ so \: from \: all \: solution \: we \: get \: the \: zeros \: are \: 6 \: \: 2 \: \: and \: 5$