Question

Factorise 12x^2 - 7x + 1 through factor theorem.

Answer 1

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Factor Theorem :

If ${p( x )}$ is a polynomial of degree ${n ≥ 1}$ and

' a ' is any real number

Then,

$(i )( x - a )$ is a factor of ${p( x ) , if\:p( a ) = 0,}$

$(ii)$ and it's converse " if ${( x - a )}$

is a factor of a polynomial ${p( x )\:then\:p ( a ) = 0},$

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Now ,

${Let\:p( x )} = 12{x}^{2} - 7x + 1,$

$=2{x}^{2} - 4x - 3x + 1$

$=4x( 3x - 1 ) - 1( 3x - 1 )$

$=( 3x - 1 )( 4x - 1 )$

$Take\:3x - 1 = 0\:or\:4x - 1 = 0$

$x =\frac{1}{3}\:or\:x = \frac{1}{4}$

$p(1/3) =12×{(1/3)}^{2} - 7 ( 1/3 ) + 1$

$=\frac{12}{9} -\frac{7}{3} + 1$

$=\frac{4}{3} -\frac{7}{3} + 1$

$=\frac{( 4 - 7 + 3 )}{3}$

$= 0$

$Therefore ,$

$x - \frac{1}{3}$ is a factor of ${p ( x )}.$

$p( 1/4 ) = 12 {( 1/4 )}^{2} - 7( 1/4)+1$

$=\frac{12}{16} -\frac{7}{4} + 1$

$=\frac{3}{4} -\frac{7}{4} + 1$

$=\frac{( 3 - 7 + 4 )}{4}$

$= 0$

${Therefore},$

$x -\frac{1}{4}$ is a factor of ${p( x )}$

Now , we conclude that ,

$\boxed{x -\frac{1}{3},x -\frac{1}{4}\: are\:two\:factors\:of\:{ p( x )}}$

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Answer 2

Answer:

Factor Theorem :

If p( x ) is a polynomial of degree n ≥ 1 and

' a ' is any real number , then

i )( x - a ) is a factor of p( x ) , if p( a ) = 0,

ii ) and it's converse " if ( x - a ) is a factor of

a polynomial p( x ) then p ( a ) = 0,

***********"***********************

Now ,

Let p( x ) = 12x² - 7x + 1 ,

= 12x² - 4x - 3x + 1

= 4x ( 3x - 1 ) - 1( 3x - 1 )

= ( 3x - 1 )( 4x - 1 )

Take 3x - 1 = 0 or 4x - 1 = 0

x = 1/3 or x = 1/4

p ( 1/3 ) = 12 × ( 1/3 )² - 7 ( 1/3 ) + 1

= 12/9 - 7/3 + 1

= 4/3 - 7/3 + 1

= ( 4 - 7 + 3 )/3

= 0

Therefore ,

x - 1/3 is a factor of p ( x ) .

2 ) p( 1/4 ) = 12 ( 1/4 )² - 7( 1/4 ) + 1

= 12/16 - 7/4 + 1

= 3/4 - 7/4 + 1

= ( 3 - 7 + 4 ) / 4

= 0

Therefore ,

x - 1/4 is a factor of p( x ).

Now , we conclude that ,

x - 1/3 , x - 1/4 are two factors of p( x ).