factorise :- 3(x²-5x)²-2(x²-5x+5)-6 step by step
Answers
Answer:
Required factorised polynomial is ( 3x^2 - 15x - 8 )( x^2 - 5x + 2 ) .
Step-by-step-explanation:
Given,
3( x^2-5x )^2 - 2( x^2-5x+5 ) - 6
Let,
x^2 - 5x = a
Now, { continued }
= > 3( a )^2 - 2( a + 5 ) - 6
= > 3a^2 - 2a - 10 - 6
= 3a^2 - 2a - 16
= > 3a^2 - ( 8 - 6 )a - 16
= > 3a^2 - 8a + 6a - 16 = 0
= > a( 3a - 8 ) + 2( 3a - 8 )
= > ( 3a - 8 )( a + 2 )
If it is not equal to 0 : ( it is a polynomial )
= > { 3( x^2 - 5x } - 8 } ( x^2 - 5x + 2 )
= > ( 3x^2 - 15x - 8 )( x^2 - 5x + 2 )
If it is equal to 0 :
= > a = 8 / 3 or - 2
Substituting the value of a :
= > x^2 - 5x = 8 / 3 Or x^2 - 5x = - 2
= > 3x^2 - 15x = 8 Or x^2 - 5x + 2 = 0
= > 3x^2 - 15x - 8 = 0 Or x^2 - 5x + 2 = 0
Using quadratic equation : If an equation is written as ax^2 + bx + c = 0 then possible values of x can be determined by using x = { - b ± √( b^2 - 4ac ) } / 2a
On the basis of the formula given above :
Case 1 : 3x^2 - 15x - 8 = 0
= > x = { 15 ± √{ 225 + 4( 8 × 3 ) } } / 2( 3 )
= > x = { 15 ± √321 } / 6
Similarly,
Case 2 : If x^2 - 5x + 2 = 0
= > x = { 5 ± √{ 25 - 4( 2 ) } / 2
= > x = ( 5 ± √17 ) / 2
Hence the required values of x are ( 5 ± √17 ) / 2 and ( 15 ± √321 ) / 6.
AND ONE MORE ANSWER
Given,
3(x² - 5x)² - 2(x² - 5x + 5) - 6
Assume ,
x² - 5x = a
Also,
= 3(a)² - 2(a + 5) - 6
= 3a² - 2a - 10 - 6
= 3a² - 2a - 16
= 3a² - (8 - 6)a - 16
3a² - 8a + 6a - 16 = 0
= a(3a - 8) + 2(3a - 8)
= (3a - 8)(a + 2)
3a - 8 = 0
3a = 8
Also,
a + 2 = 0
a = -2
If it is not equal to 0 :-
Then,
= [3(x² - 5x) - 8] (x² - 5x + 2)
= (3x² - 15x - 8)(x² - 5x + 2)
If it is equal to 0 :-
Putting the value of a :-
x² - 5x = -2
3x² - 15x = 8
x² - 5x + 2 = 0
3x² - 15x - 8 = 0
x² - 5x + 2 = 0
Quadratic Equation :
First :-
3x² - 15x - 8 = 0
Also
Second :-
If x² - 5x + 2 = 0