Math, asked by shuklashikher1, 9 months ago

factorise x^3-13x-12 use factor theorem

Answers

Answered by BrainlyNisha001

1

$\huge\mathfrak\blue{Answer:-}$

$<font color = white >$

→ p( x ) = x³ - 13x - 12

=> p( -1 ) = ( -1 )³ - 13( -1 ) -12

=> p( -1 ) = -1 + 13 - 12 = 0

=> By Remainder / Factor theorem, ( x - [ -1 ] ) is a factor

=> p( x ) = ( x + 1 )( x² - x - 12 )

• p( -3 ) = ( -3 + 1 )( 9 - ( -3 ) - 12 ) = ( -2 )( 0 ) = 0

• p( 4 ) = ( 4 + 1 )( 16 - 4 - 12 ) = 5 x 0 = 0

=> p( x ) = ( x + 1 )( x + 3 )( x - 4 )

____________________♡

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Answered by Anonymous

2

$\huge\underline \red\star \boxed{\fcolorbox{pink}{purple}{answer}}\underline \green\star$

$<font color=green>$

Let, the given polynomial be

$\green{p(x) = {x}^{3} }- \pink{ 13x - 12}$

NOW, WE WILL substitute various values of x until we get p(x)=0 as follows:

$\purple{for \: x = 0}$

$\orange{p(0) = {(0)}^{3} - 13(0) - 1}2$

$= 0 - 13 - 12$

$= -25 ≠ 0$

$.\degree.p(0)≠0$

$for \: x = 1$

$p(1) = {(1)}^{3} - 13(1) - 12$

$= 0 - 13 - 12$

$-25 ≠ 0$

$.\degree.p(1)≠0$

$for \: x = 4$

$p(4) = ( {4)}^{3} - 13(4) - 12$

$= 64 - 52 - 12$

$= 64 - 64 = 0$

$tex].\degree.p(4) = 0$ [/tex]

THUS, (x-4) is a factor of p(x)

now,

p(x)=(x-4)×g(x)...............(I)

$= > g(x) = \frac{p(x)}{(x - 4)}$

Therefore, g(x) is obtained by after dividing p(x) by (x-4) as shown attachment:

From the division, we get the quotient

$g(x) = {x}^{2} + 4x + 3$

and now we factorise it as follows:

${x}^{2} + 4x + 3$

$= {x}^{2} + x + 3x + 3$

$= x(x + 1) + 3(x + 1)$

$= (x + 3)(x + 1)$

from eq.(I), we get p(x)=(x-4)(x+3)(x+1)

$hence \: {x}^{3} - 13x - 12 = (x - 4)(x + 3)(x + 1)$

HOPE IT'S HELPS YOU ❣️

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