Question

if -2 and -6 are the zeros of polynomial x⁴+x³-34x²-4x+120 , find the other two zeroes
(don't write useless things)​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:f(x) = {x}^{4} + {x}^{3} - 34 {x}^{2} - 4x + 120$

and

$\rm :\longmapsto\: - 2 \: and \: - 6 \: are \: zeroes \: of \: f(x)$

$\rm :\implies\:(x + 2) \: and \: (x + 6) \: are \: factors \: of \: f(x)$

$\rm :\implies\:(x + 2) \: (x + 6) \: is \: factors \: of \: f(x)$

$\rm :\implies\: {x}^{2} + 8x + 12\: is \: factors \: of \: f(x)$

So, By long division we have

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\: \: \: \: \: \: \: {x}^{2} \: - 7x + \: 10 \: \: \: \: \:\:}}}\\ {{\sf{ {x}^{2} +8x + 12}}}& {\sf{\: {x}^{4} + {x}^{3} - {34x}^{2} - 4x +120\:}} \\{\sf{}}&\underline{\sf{\: \: \: { - x}^{4} - {8x}^{3} - 12{x}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:}}\\{\sf{}}&{\sf{\: \: - 7 {x}^{3} - {46x}^{2} - 4x + 120 \:\:}}\\{\sf{}}&\underline{\sf{{7x}^{3} + {56x}^{2} + 84x \: \: \: \: \:\:}}\\{\sf{}}&{\sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {10x}^{2} + 80x + 120 \:\:}}\\{\sf{}}&\underline{\sf{\:\: \: \: \: \: \: \: \: \: \: \: \: \: \: { -10x}^{3} { - 80x}- 120\:\:}}\\{\sf{}}&\underline{\sf{\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:0\:\: \: \: \: \: \: \: \: \: }}\end{array}\end{gathered}\end{gathered}\end{gathered}$

We know,

Dividend = Divisor × Quotient + Remainder

$\bf\implies \: {x}^{4} + {x}^{3} - {34x}^{2} - 4x + 120$

$\: \rm \: = \:( {x}^{2} + 8x + 12)( {x}^{2} - 7x + 10)$

$\: \rm \: = \:( {x}^{2} + 8x + 12)( {x}^{2} - 5x - 2x+ 10)$

$\: \rm \: = \:( {x}^{2} + 8x + 12)( {x}(x - 5) - 2(x - 5))$

$\: \rm \: = \:( {x}^{2} + 8x + 12)(x - 5)(x - 2)$

$\bf\implies \:Remaining \: zeroes \: of \: f(x) = 2, \: 5$

Additional Information :-

$\red{\rm :\longmapsto\: \alpha \: and \: \beta \: are \: zeroes \: of \: {ax}^{2} + bx + c \: then \:}$

$\boxed{ \sf{ \: \alpha + \beta = - \frac{b}{a}}}$

and

$\boxed{ \sf{ \: \alpha\beta = \frac{c}{a}}}$

$\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: {ax}^{3} + b {x}^{2} + cx + d \: then \:}$

$\boxed{ \sf{ \: \alpha + \beta + \gamma = - \frac{b}{a}}}$

$\boxed{ \sf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a}}}$

$\boxed{ \sf{ \: \alpha \beta \gamma = - \frac{d}{a}}}$