Question

factorize the following ​

Answer 1

EXPLANATION.

Factorizes.

⇒ 4a² - 12a - 216 = 0.

As we know that,

Take 4 as common in the equation, we get.

⇒ 4[a² - 3a - 54] = 0.

⇒ a² - 3a - 54 = 0.

Factorizes the equation into middle term splits, we get.

⇒ a² - 9a + 6a - 54 = 0.

⇒ a(a - 9) + 6(a - 9) = 0.

⇒ (a + 6)(a - 9) = 0.

⇒ a = -6  and  a = 9.

                                                                                                                       

MORE INFORMATION.

Nature of the roots of the quadratic expression.

(1) = Real and unequal, if b² - 4ac > 0.

(2) = Rational and different, if b² - 4ac is a perfect square.

(3) = Real and equal, if b² - 4ac = 0.

(4) = If D < 0 Roots are imaginary and unequal Or complex conjugate.

Answer 2

Question :

Factorise 4a² - 12a - 216

Solution :

• 4a² - 12a - 216 = 0

For this we need to take 4 as common to factorise it in a more easier way

• 4(a² - 3a - 54) = 0

• 4{a² - (9 - 6)a - 54} = 0

• 4{a² - 9a + 6a - 54} = 0

• 4{a(a - 9) + 6(a - 9)} = 0

• 4(a - 9)(a + 6) = 0

• (a - 9)(a + 6) = 0/4

• (a - 9)(a + 6) = 0

Now, there are two conditions two get the answer

1st Condition :

a - 9 = 0

• a = 9

2nd Condition :

a + 6 = 0

• a = - 6

So the answer is a = 9 or a = - 6