Question

factorise p^2+3p-18​

Answer 1

p² + 3p - 18

So we have to write 3 as a + b, where ab = - 18 × 1 = - 18.

And also it's considered that,

a = u + v

b = u - v

Now,

a + b = 3

=> u + v + u - v = 3

=> 2u = 3

=> u = 3/2

And,

ab = - 18

=> (u + v)(u - v) = - 18

=> u² - v² = - 18

=> (3/2)² - v² = - 18

=> 9/4 - v² = - 18

=> 9/4 + 18 = v²

=> (9 + 72) / 4 = v²

=> 81/4 = v²

=> v = 9/2

So,

a = u + v = 3/2 + 9/2 = (3 + 9)/2 = 12/2 = 6

b = u - v = 3/2 - 9/2 = (3 - 9)/2 = - 6/2 = -3

Hence the coefficient of x, 3, can be split as,

3 = 6 + (- 3)

Seems that there was no need to do such a huge method!!!

Okay...

p² + 3p - 18

=> p² + 6p - 3p - 18

=> p(p + 6) - 3(p + 6)

=> (p - 3)(p + 6)

Hence factorised!!!

Answer 2

Answer

here is your answer

1. = p^2 + 3p - 18
2. = p^2 + 3p - 18 = p^2 + 6p - 3p - 18
3. = p( p + 6 ) - 3 ( p + 6)
4. = (p + 6) (p - 3)

Step-by-step explanation:

In 1. we needed to find the two number that on adding give "3p" and on multiplying give "-18"

So in 2. we find those numbers i.e. 6p and -3p

In 3. we took p as common

In 4. taking ( p + 6) common we solved the question.

Hope it helps you

thank you.