Question

Find the zeroes of a polynomial x³ -5x²-16x +80, if its two zeroes are equal in magnitude but opposite in sign.

Answer 1

Hello,

To find the Zeroes we can opt for two process.
1)trial and error method

2)factorization

This polynomial can be factorized so, we will solve it by method 2 as it will be easier than the first one.

So,
$p(x) = {x}^{3} - 5 {x}^{2} - 16x + 80 \\ \\ p(x) = {x}^{3} - 16x - 5 {x}^{2} + 80 \\ = x( {x}^{2} - 16) - 5( {x}^{2} - 16) \\ \\ = ( {x}^{2} - 16)(x - 5) \\ \\ now \: first \: bracket \: can \: further \: \\ be \: factorized \\ \\ ( {x}^{2} - {4}^{2} )(x - 5) \\ = (x + 4)(x - 4)(x - 5) \\ \\ equating \: with \: zeroes \\ the \: roots \: of \: the \: p(x)are \\ 4 \: \: and \: ( - 4) \: \: and \: 5$
Therefore,
The roots which are same in magnitude but opposite in sign in 4

The roots of p(x) are

4, (-4) and 5

Hope this will be helping you

Answer 2

Answer:

→x³-5x²-16x+80

→x³-16x-5x²+80

→x(x²-16) -5(x²-16)

→(x²-16) (x-5)

(x²-16) can further be simplified:

→(x²-16) = (x²-4²)

= (x+4)(x-4) [a²-b²= (a+b)(a-b) ]

therefore, zeroes of x³-5x²-16+80 are (x+4)(x-4) (x-5)

hence zeroes are 4 , -4 and 5.