Two factoring problems that I can't figure out

Question

I have two problems I can't seem to factor.

1. $8a^3+27b^3+2a+3b$

I first tried it by grouping. I grouped the first two terms together and used sum of two cubes to get $$(2a+3b)(4a^2-6ab+9b^2)$$ and then I added the leftover $2a+3b$ to get: $$(2a+3b)(4a^2-6ab+9b^2) + 2a+3b$$ But the answer is $(2a+3b)(4a^2-6ab+9b^2 + 1)$ and I'm not quite sure how you got the $+1$ from the $2a+3b$.

2. $x^3+3x^2+3x+1$

The answer is $(x+1)^3$ but I don't know how to get there. I tried factoring by grouping, but that wasn't right, and I'm stuck as to how to solve it. Thanks!

Answer 1

.For the first question, you were very close to the answer. In fact, when you write $(2a+3b)(4a^2+9b^2-6ab) + 2a+3b$, all you need to notice is that the $2a+3b$ at the end can also be taken into the bracket, using the distributive law: $$ (2a+3b)(4a^2+9b^2-6ab) + 2a+3b = (2a+3b)(4a^2+9b^2-6ab) + 1(2a+3b) \\ = (2a+3b)(4a^2+9b^2-6ab + 1) $$ Using $xy+ xz = x(y+z)$, where $x=2a+3b, y = 4a^2+9b^2-6ab,z=1$.

Hence the answer follows.

For the second one, yes it is tricky, but here is the catch, or rather a heuristic in these kind of problems:

Suppose you are asked to factorize some complicated cubic, say $x^3+ax^2+bx+c$, or any other complicated expression. Then, most likely, the roots are integers, which are factors of the constant in the expression.

For example, here $x^3+3x^2+3x+1$ is given, and the constant is $1$. Hence, the roots, if they are integers, are all factors of $1$. Then, they can only be $-1$ or $1$. You can use the remainder theorem to check which of these is a factor, by substituting $-1$ and $1$ in the expression and checking when the value is zero. Turns out for $-1$ that the expression is $0$, so $(x-(-1)) = (x+1)$ is a factor of your expression. Then, go on and divide by $1$, you will get $x^2+2x+1$, which is $(x+1)^2$. So collecting everything, we have $x^3+3x^2+3x+1=(x+1)^3$.

It's good to do one more example. Suppose someone asks you to factor $x^3-x^2-4x+4$. Using the heuristic, you know that the roots must divide $4$. Hence, the roots can only be $\pm 1,\pm 2,\pm4$. Use remainder theorem, and check that substituting $1$ gives $0$. Hence $x-1$ is a factor, and on dividing you get $x^2-4$, which is $(x-2)(x+2)$. Hence the complete answer is $(x-1)(x+2)(x-2)$.

Answer 2

For the first one you are using the distributive law. You have $(2a+3b)(4a^2-6ab+9b^2) + (2a+3b)\cdot 1$ and can combine the right terms that are multiplied by $(2a+3b)$

For the second, I think you are expected to recognize the coefficients as a row of Pascal's triangle. Alternately you can use the rational root theorem. The rational roots of $x^3+3x^2+3x+1$ can only be $\pm 1$, so try dividing in $x+1$ and $x-1$. You come out with no remainder with $x+1$ and are left with a quadratic. Yet a third way is to see $x^3+1$ as a sum of cubes, factoring as $(x+1)(x^2-x+1)$ and factor $3x^2+3x=3x(x+1)$ and again use the distributive law to combine the two terms.

Answer 3

$$\text{2nd problem}$$ $$x^3−3x^2+3x−1$$ $$=x^3−x^2−2x^2+3x−1$$ $$=x^2(x−1)−2x^2+2x+x−1$$ $$=x^2(x−1)−2x(x−1)+1(x−1)$$ $$=(x−1)(x2−2x+1)$$ $$=(x−1)(x−1)^2$$ $$=(x−1)^3$$