Question

Find the zeroes of the polynomial :

f(x) = x³ + 3x² - 6x - 8

Please Help. Emergency​

Answer 1

Given to find the zeros of the polynomial:-

f(x) = x³ + 3x² -6x - 8

Explanation:-

Lets find the zeros by synthetic division method.

To solve in this method First we need to get the one zero. To get one zero Lets do trial and error method.

f(x) = x³ + 3x² -6x -8

x = 1

⇒f(1) = (1)³ + 3(1)² -6(1) -8

⇒f(1) = 1 + 3 -6-8

⇒f(1) = 4-14

⇒f(1) = -10

Hence it is not a zero .

x = -1

⇒f(-1) = (-1)³ + 3(-1)² -6(-1) -8

⇒f(-1) = -1 +3 +6-8

⇒f(-1) = -9+9

⇒f(-1) = 0

Hence, -1 is one zero.

Now solving by synthetic division.

Writing the Quadratic polynomial from that :-

⇒x² + 2x -8

Factorising the polynomial.

⇒x² + 4x -2x -8

⇒x(x+4) -2(x+4)

⇒(x+4)(x-2)

⇒x+4=0

⇒x = -4

⇒x-2 =0

⇒x=2

So, the value of x are -1, 2, -4 .

So, the zeros of given cubic polynomial are 1, -2, -4 .

${}$

Answer 2

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

Given cubic polynomial is

$\rm :\longmapsto\:f(x) = {x}^{3} + {3x}^{2} - 6x - 8$

Since, it's a cubic polynomial, so as to factorize cubic polynomial, we use hit and trial method to find the one zero of cubic polynomial.

Here, constant term is 8, so factors of 8 are

$\rm :\longmapsto\: \pm \: 1, \: \pm \:2, \: \pm \:4, \: \pm \:8$

So, Let consider

$\rm :\longmapsto\:f( - 1) = {( - 1)}^{3} + {3( - 1)}^{2} - 6( - 1) - 8$

$\rm :\longmapsto\:f( - 1) = - 1 + 3 + 6 - 8$

$\rm :\longmapsto\:f( - 1) = - 9 + 9$

$\rm :\longmapsto\:f( - 1) = 0$

$\bf\implies \:x + 1 \: is \: factor \: of \: f(x)$

Now, using Long Division Method, we have

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\:\: {x}^{2} + 2 x - 8\:\:}}}\\ {\underline{\sf{x + 1}}}& {\sf{\: {x}^{3} + {3x}^{2} - 6x - 8 \:\:}} \\{\sf{}}& \underline{\sf{ -  {x}^{3} - {x}^{2}   \: \: \: \: \: \: \: \: \: \: \: \: \: \:  \:  \: \:\:}} \\ {{\sf{}}}& {\sf{\: \:  \: \: \: \:2{x}^{2} - 6x - 8  }} \\{\sf{}}& \underline{\sf{\:\:    \:  \:  - 2 {x}^{2}  - 2x  \:  \:  \:  \:  \:  \: \:\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \:  \:  \:  \:  \:  \:  \:    - 8x - 8  \:\:}} \\{\sf{}}& \underline{\sf{\: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:8x + 8\:\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \:  \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0\:\:}}  \end{array}\end{gathered}\end{gathered}\end{gathered}$

So, By Division Algorithm, we have

$\rm :\longmapsto\:f(x) = {x}^{3} + {3x}^{2} - 6x - 8$

$\rm \:  =  \: (x + 1)( {x}^{2} + 2x - 8)$

$\rm \:  =  \: (x + 1)( {x}^{2} + 4x - 2x - 8)$

$\rm \:  =  \: (x + 1)[x(x + 4) - 2(x + 4)]$

$\rm \:  =  \: (x + 1)(x + 4)(x - 2)$

Hence,

$\rm\implies \:\boxed{\tt{ {x}^{3} + {3x}^{2} - 6x - 8 = (x + 4)(x + 1)(x - 2)}}$

Hence,

Zeroes of f(x) are - 4, 2, - 1

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More Identities to know :

➢  (a + b)² = a² + 2ab + b²

➢  (a - b)² = a² - 2ab + b²

➢  a² - b² = (a + b)(a - b)

➢  (a + b)² = (a - b)² + 4ab

➢  (a - b)² = (a + b)² - 4ab

➢  (a + b)² + (a - b)² = 2(a² + b²)

➢  (a + b)³ = a³ + b³ + 3ab(a + b)

➢  (a - b)³ = a³ - b³ - 3ab(a - b)