Question

Given P(x)=x^3-4x^2+4x-16, find the zeros, real and non-real of P
The zeros are:
Write P in factored form (as a product of linear factors). Be sure to write the full equation, including P(x)=.

Answer 1

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

Given cubic polynomial is

$\rm :\longmapsto\:P(x) = {x}^{3} - {4x}^{2} + 4x - 16$

$\rm :\longmapsto\:P(x) = {x}^{2}(x - 4) + 4(x - 4)$

$\rm :\longmapsto\:P(x) = ({x}^{2} + 4)(x - 4)$

can be further rewritten as

$\rm :\longmapsto\:P(x) = ({x}^{2} - ( - 4))(x - 4)$

$\rm :\longmapsto\:P(x) = ({x}^{2} - 4 {i}^{2} )(x - 4)$

$\red{ \bigg\{  \sf \: \because \: {i}^{2} = - 1 \bigg\}}$

$\rm :\longmapsto\:P(x) = ({x}^{2} - {(2i)}^{2} )(x - 4)$

$\rm :\longmapsto\:P(x) = (x + 2i)(x - 2i)(x - 4)$

$\red{ \bigg\{  \sf \: \because \: {x}^{2} - {y}^{2} = (x - y)(x + y) \bigg\}}$

So, zeroes of the polynomial P(x) are

$\rm :\longmapsto\:x = 2i \: \: or \: \: - 2i \: \: or \: \: 4$

Hence,

Factorized form of P(x)

$\green{\rm :\longmapsto\:\boxed{\tt{ P(x) = (x + 2i)(x - 2i)(x - 4)}}}$

And zeroes of P(x) are

$\purple{\rm :\longmapsto\:\boxed{\tt{ 4, \: \: 2i, \: \: - 2i}}}$

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Additional Information

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

$\boxed{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{b}{a}}}$

$\boxed{ \bf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}}}$

$\boxed{ \bf{ \: \alpha \beta \gamma = - \dfrac{d}{a}}}$

Answer 2

$\color{blue}{ \colorbox{orange}{\colorbox{red} {hope \: it \: helps \: you}}}$