Question

Show that (3x + 2) is a factor of (6x3 + 13x2 - 4) and hence factorise (6x3 + 13x2
4).​

Answer 1

Answer:

By factor theorum , we know that

If f ( x ) is a factor of p ( x ) , then p ( X ) = 0

Hence , lets put 3x + 2 = 0

3x = -2

x = - 2 / 3

Putting x = - 2 / 3 in given Expression , we get ;

= 6 * ( - 2 / 3 )³ + 13 ( - 2 / 3 )² - 4

= ( -16 / 9 ) + ( 52 / 9 ) - 4

= ( 52 - 16 ) / 9 - 4

= ( 36 / 9 ) - 4

= ( 4 - 4 ) = 0

Hence , when x = -2 / 3 , then p ( x ) = 0

Hence , by factor theorum , it justifies that ( 3x + 2 ) is a factor of ( 6x³ + 13x² - 4 ) .

Now ,

Divide p (x) by f (x) ,

( 6x³ + 13x² - 4 ) / ( 3x + 2 ) = 2x² + 3x - 2

( Division is given in attachment )

Now ,

By the method of splitting middle term , factorize 2x² + 3x - 2 ,

=) 2x² + 3x - 2

=) 2x² + 4x - x - 2

=) 2x ( x + 2 ) - 1 ( x + 2 )

=) ( 2x -1 )( x + 2 )

=) x = 1 / 2 , x = -2

Now ,

Factor 1st :

x = 1 / 2

x - ( 1 / 2 ) = 0

2x - 1 = 0 ( multiplying both side by 2 )

Factor 2nd :

x = -2

x + 2 = 0

Factor 3rd :

x = - 2 / 3

x + ( 2 / 3 ) = 0.

3x + 2 = 0 ( multiplying both side by 3 )

Hence , we got that , ( 3x + 2 ) , ( x +2 ) & ( 2x -1 ) are the factors of 6x³ + 13x² - 4 .

Hence ,

6x³ + 13x² - 4 = ( 3x + 2 )( x + 2 )( 2x - 1 )

Hence , factorized .

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Answer 2

Please mark this answer as the brainliest

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

This is the quotient of long division