# Factorise b^2 - 21b + 108

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### 3 Answers

We could also apply quadratic formula:

b1 = [-(-21)+sqrt((-21)^2 + 4*108)]/2*1

b1 = (21+sqrt9)/2

b1 = (21+3)/2

b1 = 12

b2 = (21-3)/2

b2 = 9

We can write the quadratic expression as a product of linear factors:

b^2 - 21b + 108 = (b - b1)(b - b2)

b^2 - 21b + 108 = (b - 12)(b - 9)

Remark: b1 and b2 are the roots of the quadratic equation:

b^2 - 21b + 108 = 0

To factorise b^2-21b +108.

We do the factorisatio by gruouping method.

We split the middle term -21b in two in such a way that their product is equal to the product of end terms b^2 and 108, or equal to 108b^2.

Therefore middle term -21b = -12b -9b and

(-12b)(-9b) = 108b^2.

Therefore b^2-21b +108 = b^2-12b-9b+108

b^2-21b +108 = b(b-12)-9(b-12)

b^2-21b +108 = (b-12)(b-9).

Therefore the factor form of the given expression is b^2-21b+108 = (b-12)(b-9).

` b^2 - 21b + 108 `

a=1 b=-21 c=108

use axxc then find factors of 108 that add to -21

`1xx108=108 `

factors of 108 that = -21 are -9 and -12

plug them in

`b^2 - 9b-12b + 108 `

factor

`(b^2 - 9b)(-12b + 108) `

`b(b-9)-12(b-9) `

`(b-12)(b-9) `

b=12 b=9