Factoring trinomials

[Gr8fu13]

Directions: Factor the trinomial
b^3 - 3b^2 - 40b

I tried to factor and came up with this:
Substitute u = b.
u^2 - 3u - 40
u(u + 5)(u - 8)
u^2 + 5u - 8u - 40
u^2 - 3u - 40
I put the original x=b back into the answer.
b^2(b + 5)(b - 8)
b^3 + 5b^2 - 8b^2 - 40b
b^3 - 3b^2 - 40b
So would this be the answer?
b^2(b + 5)(b - 8)

I always get confused on these problems

 

[JeffM]

Directions: Factor the trinomial
b^3 - 3b^2 - 40b

I tried to factor and came up with this:
Substitute u = b.
u^2 - 3u - 40
u(u + 5)(u - 8)
u^2 + 5u - 8u - 40
u^2 - 3u - 40
I put the original x=b back into the answer.
b^2(b + 5)(b - 8)
b^3 + 5b^2 - 8b^2 - 40b
b^3 - 3b^2 - 40b
So would this be the answer?
b^2(b + 5)(b - 8)

I always get confused on these problems

If you substitute u = b, is not the revised expression just u^3 - 3u^2 - 40u, which is not different in any important respect from the original expression. You cannot go dropping exponents just because you have changed your notation. A rose by any other name would smell as sweet, and changing from b to u does not change the problem at all.

OK When you factor a polynomial, you are looking for a set of expressions that when multiplied give you the polynomial. Do you see ANYTHING that is a common factor in each summand of the polynomial?

In general, every polynomial in one variable with no constant term has an IMMEDIATE preliminary factorization that can be done. Do you see it?

au
au^2 + bu
au^3 + bu^2 + cu
au^4 + bu^3 + cu^2 + du.

Each polynomial above has a common factor, which is?????

 

[Denis]

Directions: Factor the trinomial
b^3 - 3b^2 - 40b
I tried to factor and came up with this:
Substitute u = b.

No need to substitute; you seem quite confused with "substituting"; better get that straight...
b is common to all terms, so:
b(b^2 - 3b - 40)
then factor:
b(b - 8)(b + 5)

Done.