Find the square root of(6x²+5x-6)(6x²-x-2)(4x²+8x+3)
Answers
Answer:
12x³ + 16x² - 7x - 6
Step-by-step explanation:
whole root of ( 6x² + 5x - 6 ) ( 6x² - x - 2 ) ( 4x² + 8x + 3 )
( 6x² + 5x - 6 )
= ( 6x² + 9x - 4x - 6 )
= 3x ( 2x + 3 ) - 2 ( 2x + 3 )
= ( 2x + 3 ) ( 3x - 2 ) ------> 1.
( 6x² - x - 2 )
= ( 6x² + 3x - 4x - 2 )
= 3x ( 2x + 1 ) - 2 ( 2x + 1 )
= ( 2x + 1 ) ( 3x - 2 ) ------> 2.
( 4x² + 8x + 3 )
= ( 4x² + 2x + 6x + 3 )
= 2x ( 2x + 1 ) + 3 ( 2x + 1 )
= ( 2x + 1 ) ( 2x + 3 ) ------> 3.
whole root of ( 6x² + 5x - 6 ) ( 6x² - x - 2 ) ( 4x² + 8x + 3 ),
Substitute 1, 2, 3
=> whole root of
( 2x + 3 ) ( 3x - 2 ) ( 2x + 1 ) ( 3x - 2 ) ( 2x + 1 ) ( 2x + 3 )
=> whole root of
( 2x + 3 ) ( 2x + 3 ) ( 3x - 2 ) ( 3x - 2 ) ( 2x + 1 ) ( 2x + 1 )
=> whole root of
( 2x + 3 )² × ( 3x - 2 )² × ( 2x + 1 )²
=> ( 2x + 3 ) ( 3x - 2 ) ( 2x + 1 )
=> [ ( 2x + 3 ) ( 3x - 2 ) ] ( 2x + 1 )
=> [ 2x ( 3x - 2 ) + 3 ( 3x - 2 ) ] ( 2x + 1 )
=> [ 2x ( 3x ) + 2x ( - 2 ) + 3 ( 3x ) + 3 ( - 2 ) ] ( 2x + 1 )
=> [ 6x² - 4x + 9x - 6 ] ( 2x + 1 )
=> [ 6x² + 5x - 6 ] ( 2x + 1 )
=> ( 2x + 1 ) [ 6x² + 5x - 6 ]
=> 2x ( 6x² + 5x - 6 ) + 1 ( 6x² + 5x - 6 )
=> 2x ( 6x² ) + 2x ( 5x ) + 2x ( - 6 ) + 6x² + 5x - 6
=> 12x³ + 10x² - 12x + 6x² + 5x - 6
=> 12x³ + 10x² + 6x² - 12x + 5x - 6
=> 12x³ + 16x² - 7x - 6
( It look like simple problem, but more steps are there to find the answer. You should be careful while, doing this kind of problems. I have taken 30 minutes to find the answer and then type it. I have done my best to do this sum. I hope this helps you. )