Question

factorise y^3-6y^2-19y+84​

Answer 1

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

Given expression is

$\rm :\longmapsto\: {y}^{3} - {6y}^{2} - 19y + 84$

Let assume that

$\rm :\longmapsto\: f(y) = {y}^{3} - {6y}^{2} - 19y + 84$

Now, we have to find the zero of f(y) by hit and trial method.

Let y = 1

$\rm :\longmapsto\: f(1) = {1}^{3} - {6(1)}^{2} - 19(1) + 84$

$\rm :\longmapsto\: f(1) = 1 - 6 - 19 + 84 = 60 \ne \: 0$

Let y = 2

$\rm :\longmapsto\: f(2) = {2}^{3} - {6(2)}^{2} - 19(2) + 84$

$\rm :\longmapsto\: f(2) = 8 - 24 - 38 + 84 = 30 \ne \: 0$

Let y = 3

$\rm :\longmapsto\: f(3) = {3}^{3} - {6(3)}^{2} - 19(3) + 84$

$\rm :\longmapsto\: f(3) =27 - 54 - 57 + 84 = 111 - 111 = 0$

$\bf\implies \:3 \: is \: the \: zero \: of \: f(y)$

$\bf\implies \:y - 3 \: is \: the \: factor \: of \: f(y)$

So, using long division, we have

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\:\: {y}^{2} - 3y - 28\:\:}}}\\ {\underline{\sf{y - 3}}}& {\sf{\: {y}^{3}  -  {6y}^{2} - 19y + 84 \:\:}} \\{\sf{}}& \underline{\sf{- {y}^{3} + 3 {y}^{2}   \:  \:  \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\:}} \\ {{\sf{}}}& {\sf{\: \: \: \: \: \: \: \: \: \: \: \: - 3{y}^{2} -19y + 84 \:  \:\:}} \\{\sf{}}& \underline{\sf{\:\: \:  \:  \:    \:  \:  \:  3{y}^{2}  - 9y  \:  \:  \:  \:  \:  \: \:\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \:  \:  \:  \:  \:  \: \: \: \: \: \: - 28y + 84  \:\:}} \\{\sf{}}& \underline{\sf{\: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  28y + 84\:\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \:  \: \:  \:  \:  \:  \:  \:  \:  \: 0\:\:}}  \end{array}\end{gathered}\end{gathered}\end{gathered}$

We know,

Dividend = Divisor × Quotient + Remainder

Therefore,

$\rm :\longmapsto\: {y}^{3} - {6y}^{2} - 19y + 84$

$\rm \: = \:(y - 3)( {y}^{2} - 3y - 28)$

$\rm \: = \:(y - 3)( {y}^{2} - 7y + 4y - 28)$

$\rm \: = \:(y - 3)\bigg(y(y - 7) + 4(y - 7)\bigg)$

$\rm \: = \:(y - 3)(y - 7)(y + 4)$

Hence,

$\boxed{ \bf{ \: \: {y}^{3} - {6y}^{2} - 19y + 84 = (y - 3)(y + 4)(y - 7)}}$

Additional Information :-

$\boxed{ \bf{ \: {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2}}}$

$\boxed{ \bf{ \: {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2}}}$

$\boxed{ \bf{ \: {(x + y)}^{3} = {x}^{3} + 3xy(x + y) + {y}^{3}}}$

$\boxed{ \bf{ \: {(x - y)}^{3} = {x}^{3} - 3xy(x - y) - {y}^{3}}}$

$\boxed{ \bf{ \: {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2})}}$

$\boxed{ \bf{ \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2})}}$

$\boxed{ \bf{ \: {x}^{4} - {y}^{4} = (x - y)(x + y)( {x}^{2} + {y}^{2})}}$

$\boxed{ \bf{ \: {x}^{2} - {y}^{2} = (x - y)(x + y)}}$