Can You Factor This?

Question

8a^3 + 1

2x^3 - 6x^2 + x - 3

tw - 3w - w^2 + 3t

9p^2 - 4q^2

Answer 1

1) Difference of cubes.

2) Grouping.

3) Rearranging and then grouping.

4) Difference of squares.

Answer 2

Answer 1:
8a^3 + 1
= (2a)^3 + (1)^3
= (2a + 1)(4a^2 + 1 + 2a)

Answer 2:
2x^3 - 6x^2 + x - 3
Let p(x) = 2x^3 - 6x^2 + x - 3
Put x = 3

p(3) = 2(3)^3 - 6(3)^2 + 3 - 3
= 54 - 54 + 3 - 3
= 0
3 is a zero of p(x)
So, (x - 3) is a factor of p(x)

By synthetic division,
p(x) = (x - 3)(2x^2 + 1)

Answer 3:
tw - 3w - w^2 + 3t
= -w^2 - 2w + 3t
= w(-2 - w) + 3t

Answer 4:
9p^2 - 4q^2
= (3p)^2 - (2q)^2
= (3p + 2q)(3p - 2q)

Answer 3

1. 8a^3 + 1 = (2a^3) + 1^3 = (2a + 1)(4a^2 - 2a + 1) =
= 8(a + 1/2)(a^2 - a/2 + 1/4)

[Use the identity: a^3+b^3 = (a+b)(a^2 - ab + b^2)]

2. Just saw that x=3 is a root:
Let's divide the whole polynomial by (x-3)

Subtracting 2x^2(x-3) = 2x^2 - 6x^2 from 2x^3 - 6x^2 + x - 3 we get
x - 3 = 1* (x-3)

Thus 2x^3 - 6x^2 + x - 3 = (x-3)(2x^2 + 1) = 2(x-3)(x^2 + 1/2)




3. Let's solve tw-3w-w^2 + 3t =0 for a constant t:

-w^2 + (t-3)w + 3t = 0

w1,2 = [(3-t) +- sqrt[(3-t)^2 + 12t]]/-2 =
= [(3-t) +- sqrt(9 - 6t + t^2 + 12t)]/-2 =
= [(3-t) +- sqrt(9 + 6t + t^2)]/ -2 =
[(3-t) +-sqrt(t+3)^2]/ -2

w1 = -3
w2 = t

-(w-t)(w+3)

4) 9p^2 = (3p)^2, 4q^2 = (2q)^2

9p^2-4q^2 = (3p)^2 - (2q)^2 = (3p -2q)(3p + 2q)

Answer 4

8a^3 + 1=(2a+1)([2a^2]+2a+1)
2x^3 - 6x^2 + x - 3=[2x^2](x-3)+x-3=(x-3)(2x^2+1)
tw - 3w - w^2 + 3t=t(w+3)-w(3+w)=(t-w)(3+w)
9p^2 - 4q^2=(3p-2q)(3p+2q)

Answer 5

Why wouldn't kids complete their homeworks themselves?

Answer 6

(2a+1)(4a^2-2a+1)
(x-3)(2x^2+1)
(w-3)(t-3)
(3p-2q)(3p+2q)

Answer 7

8a^3+1
=(2a)^2+1^2
=(2a+1)^3-3.2a.1(2a+1)
=(2a+1){(2a+1)^2-3.2a.1}
=(2a+1)(4a^2+4a+1-6a)
=(2a+1)(4a^2-2a+1)...........................ans


2x^3-6x^2+x-3
=2x^2(x-3)+1(x-3)
=(x-3)(2x^2+1).............................ans


tw-3w-w^2+3t
=tw+3t-w^2-3w
=t(w-3)-3w(w-3)
=(w-3)(t-3)..........................ans


9p^2-4q^2
=(3p)^2-(2q)^2
=(3p+2q)(3p-2q)...............................ans