plese answer all factorisation questions
Answers
EXPLANATION.
(1) = x² - 2x - 15.
As we know that,
Factorizes the equation into middle term splits, we get.
⇒ x² - 5x + 3x - 15 = 0.
⇒ x(x - 5) + 3(x - 5) = 0.
⇒ (x + 3)(x - 5) = 0.
⇒ x = - 3 and x = 5.
(2) = x² - 1x - 5.
As we know that,
⇒ D = Discriminant Or b² - 4ac.
⇒ D = (-1)² - 4(1)(-5).
⇒ D = 1 + 20.
⇒ D = 21.
⇒ α = - b + √D/2a.
⇒ α = -(-1) + √21/2.
⇒ α = 1 + √21/2.
⇒ β = - b - √D/2a.
⇒ β = -(-1) - √21/2.
⇒ β = 1 - √21/2.
(3) = x² - 7x + 12.
As we know that,
Factorizes the equation into middle term splits, we get.
⇒ x² - 4x - 3x + 12 = 0.
⇒ x(x - 4) - 3(x - 4) = 0.
⇒ (x - 3)(x - 4) = 0.
⇒ x = 3 and x = 4.
(4) = x² - 1x - 56.
As we know that,
Factorizes the equation into middle term splits, we get.
⇒ x² - 8x + 7x - 56 = 0.
⇒ x(x - 8) + 7(x - 8) = 0.
⇒ (x + 7)(x - 8) = 0.
⇒ x = -7 and x = 8.
(5) = x² + 8x + 12.
As we know that,
Factorizes the equation into middle term splits, we get.
⇒ x² + 6x + 2x + 12 = 0.
⇒ x(x + 6) + 2(x + 6) = 0.
⇒ (x + 2)(x + 6) = 0.
⇒ x = -2 and x = -6.
(6) = x² + 1x - 5.
As we know that,
⇒ D = Discriminant Or b² - 4ac.
⇒ D = (1)² - 4(1)(-5).
⇒ D = 1 + 20.
⇒ D = 21.
⇒ α = - b + √D/2a.
⇒ α = -1 + √21/2.
⇒ β = - b - √D/2a.
⇒ β = -1 - √21/2.
(7) = x² + 9x + 20.
As we know that,
Factorizes the equation into middle term splits, we get.
⇒ x² + 5x + 4x + 20 = 0.
⇒ x(x + 5) + 4(x + 5) = 0.
⇒ (x + 4)(x + 5) = 0.
⇒ x = - 4 and x = -5.
(8) = x² - 2x - 63.
As we know that,
Factorizes the equation into middle term splits, we get.
⇒ x² - 9x + 7x - 63 = 0.
⇒ x(x - 9) + 7(x - 9) = 0.
⇒ (x + 7)(x - 9) = 0.
⇒ x = -7 and x = 9.
To Factorize :-
i) x² - 2x - 15
ii) x² - 1x - 5
iii) x² - 7x + 12
iv) x² - 1x - 56
v) x² + 8x + 12
vi) x² + 1x - 5
vii) x² + 9x + 20
viii) x² - 2x - 63
Used Concepts :-
For ax² + bx + c :-
- We have to split the middle term of ax² + bx + c as ax² + nx + vx + c note that nx + vx = bx and ax² × c = nx × vx must .
Solution :-
i) x² - 2x - 15
=> x² - 5x + 3x - 15
=> x ( x - 5 ) + 3 ( x - 5 )
=> ( x - 5 ) ( x + 3 )
ii) x² - 1x - 5 = x² - x - 5
=> We can't do this question with factorization method . So , lets do this with Quadratic Formula ,
Here , a = 1 , b = -1 , c = -5
=> D = b² - 4ac = ( -1 )² - 4 × 1 × -5 = 1 + 20 = 21
=> √D = √21
So , x² - x - 5 can be Factorized into :-
=> ( -b + √D/2a ) ( -b - √D/2a )
=> ( - ( -1 ) + √21/2 × 1 ) ( - ( -1 ) - √21/2 × 1 )
=> ( 1 + √21/2 ) ( 1 - √21/2 )
iii) x² - 7x + 12
=> x² - 4x - 3x + 12
=> x ( x - 4 ) -3 ( x - 4 )
=> ( x - 4 ) ( x - 3 )
iv) x² - 1x - 56 = x² - x - 56
=> x² - 8x + 7x - 56
=> x ( x - 8 ) + 7 ( x - 8 )
=> ( x - 8 ) ( x + 7 )
v) x² + 8x + 12
=> x² + 6x + 2x + 12
=> x ( x + 6 ) + 2 ( x + 6 )
=> ( x + 6 ) ( x + 2 )
vi) x² + 1x - 5 = x² + x - 5
=> Again , we can't do this by factorization method . So lets do this by Quadratic Formulae ,
Here ,
=> a = 1 , b = 1 , c = -5
=> D = b² - 4ac = ( 1 )² - 4 × 1 × -5 = 1 + 20 = 21
=> √D = √21
So , x² + x - 5 can be Factorized into :-
=> ( -b + √D/2a ) ( -b - √D/2a )
=> ( -1 + √21/2 × 1 ) ( -1 - √21/2 × 1 )
=> ( -1 + √21/2 ) ( -1 - √21/2 )
vii) x² + 9x + 20
=> x² + 4x + 5x + 20
=> x ( x + 4 ) + 5 ( x + 4 )
=> ( x + 4 ) ( x + 5 )
viii) x² - 2x - 63
=> x² + 7x - 9x - 63
=> x ( x + 7 ) -9 ( x + 7 )
=> ( x + 7 ) ( x - 9 )
Note :-
There are some Quadratic equations which we can't do by completing the square method and factorization but Quadratic Formulae is always applicable ( सदाबहार ) .