Question

Factorise... 9a^2 + 4b^2 +6a+ 1

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Answer 1

${9a}^{2} + {4b}^{2} + 6a + 1 \\ \\ ({9a}^{2} + 6a + 1 )+ {4b}^{2}$

Factorise The Term

9a²+6a+1 By Middle Term Splitting.

$( {9a}^{2} + 6 a+ 1) + {4b}^{2} \\ \\ {9a}^{2} + 3a + 3a + 1 \\ \\ 3a(3a + 1) + 1(3a + 1) + {4b}^{2} \\ \\ (3a + 1)(3a + 1) + {4b}^{2}$

In Middle Term Splitting We Split Middle Term Such That on Multiplying it becomes Product of First and Last Term of Polynomial.

Like Here 9a²+6a+1

Middle Term=6a

can written as 3a+3a On Multiplying it it b come 9a² Which is Product of First and Last Term.

$(3a + 1)(3a + 1) + {4b}^{2}$

(3a+1)²+(2b)²

(3a+1)²+(2b)²

Answer 2

Answer:

Step-by-step explanation:

Given :

To factorise 9a² - 4b² + 6a + 1

Solution :

9a² - 4b² + 6a + 1

⇒ 9a² + 6a + 1 - 4b²

⇒ 9a² + 3a + 3a + 1 - 4b²

⇒ 3a (3a + 1) + 1 (3a + 1) - 2b (2b)

⇒ (3a + 1) (3a + 1) - (2b)²

⇒ (3a + 1)² - (2b)²

We know that,

a² - b² = (a + b) (a - b)

⇒ (3a + 1)² - (2b)²

⇒ [(3a + 1) + 2b] [(3a + 1) - 2b]

⇒ (3a + 2b + 1) (3a - 2b + 1)

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If,

9a² + 4b² +6a+ 1 is the real question,.

We know that,

i = imaginary number = √-1

⇒ i² = -1

⇒ 9a² + 4b² + 6a + 1

⇒ 9a² + 6a + 1 + 4b²

⇒ (3a + 1)² - 4i²b²

⇒ (3a + 2ib + 1) (3a - 2ib + 1)