Question

Extra Credit Problems?
Will fan and medal!
1: Factor completely 5x(x + 3) + 6(x + 3).
A. (x + 3)(30x)
B. (x + 3)(5x + 6)
C. (x + 3)(11x)
D. prime

2: Factor completely 4ab − 7ax + 8b − 14x.
A. (4b − 7x)(a + 2)
B. (4b + 7x)(a + 2)
C. (4b − 7x)(a − 2)
D.prime

3: Factor completely 9xy + 36x − 10y − 40.
A. prime
B. (y − 4)(9x − 10)
C. (y + 4)(9x + 10)
D. (y + 4)(9x − 10)

Answer 1

I won't answer them, that's your task... I won't get the extra credit... you will

but I can explain how it works...

in the 1st question what is the obvious common factor...?

Answer 2

On the first one, Try doing the common factor, with that, I mean the factor that is repeated on both terms.
As an example:
\[a^2(l+5+b^2)-b^2(l+5+b^2)-(l+5+b^2)\]
If I multiply and divide by (l+5+b^2) I get this:
\[\frac{ (l+5+b^2)(a^2(l+5+b^2)-b^2(l+5+b^2)-(l+5+b^2)) }{ (l+5+b^2) }\]
And doing some very simple algebra:
\[(l+5+b^2)(\frac{ a^2(l+5+b^2) }{ (l+5+b^2) }-\frac{ b^2(l+5+b^2) }{ (l+5+b^2) }+\frac{ (l+5+b^2) }{ (l+5+b^2) })\]
And operating the denominators with the numerators:
\[(l+5+b^2)(a^2-b^2+1)\]

That would be The "long way" to apply the "common factor" axiom.