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Q: What is the factor of b2 plus 12b plus 32?

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(a + b)(b + c)

It cannot factor without imaginary numbers

b2(b + 6)

4(16a4 + b2)

a2 - 4a + 4

(a+b+c) (a+b-c)

a(b+3)+b(b+3)

the answer to factorising (a x a3 + 2ab + b2) would be (a4+2ab+b2)

(a + b)(b - 2c)

(a + b)(b - 2c)

(b + 2)(b + 2) or (b + 2)2

as a2+b2=(a+ib)(a-ib), b2+2=[b - i sqrt(2)] [b + i sqrt(2)]

(-8 + b2) - (5 + b2) = -8 + b2 - 5 - b2 = -13

ab-2ac+b^2-2bc

b2+14b+49 = (b+7)(b+7) when factored

6(b - ac + b2 - bc)

6(ab - ac + b2 - bc)

It cannot be factored because the discriminant of b2-4ac is less than zero.

a2-b2 = (a-b)(a+b)

a2+2a2b+2ab2+b2

4a2 + 4ab - y2 + b2 You cannot simplify this expression because it does not contain like terms, but you can factor it such as: 4a2 + 4ab - y2 + b2 = (4a2 + 4ab + b2) - y2 = (2a + b)2 - y2 = [(2a + b) - y][(2a + b) + y]

b2 + 16b + 64 can be factorized to (b+8)(b+8). If they were equal to zero as in b2 + 16b + 64=0 then b = -8.

24a + b2 + 3a + 2b2= 27a + 3b2

9a4-b2=36a-2b=2(18-b) But if you meant 9a^4-b^2 it is not possible to factor

a2