asimon2005 said:
1) 2x^2 + 13x + 15
2) 2x^2 + 11x + 15
I really do need help on these please help
Hmmm....it looks like you really want us to do all of these problems for you. But, that wouldn't help you learn how to do them yourself!
I'll show you a method that will work on any
factorable trinomial of the form
ax<SUP>2</SUP> + bx + c
I'll use the above problem to illustrate.
Step 1: Multiply "a" times "c" (or, the coefficient of the x<SUP>2</SUP> times the constant term.)
For 2x<SUP>2</SUP> + 13x + 15, that means we multiply 2*15 to get 30.
Step 2: Look for two numbers which
multiply to +30, and which
add up to the coefficient of the middle term, which is +13. We can use 10 and 3, since 10*3 is 30, and 10 + 3 is 13.
Use these numbers to rewrite the middle term. Instead of 13x, use 10x + 3x:
2x<SUP>2</SUP> + 10x + 3x + 15
IF THERE ARE NO two factors of the product "ac" which add up to the middle coefficient, b, then you know your trinomial cannot be factored over the integers, and you can stop here.
Step 3: Factor by grouping. Group the first two terms together, and the last two terms together:
2x<SUP>2</SUP> + 10x + 3x + 15
Remove a common factor of
2x from the first two terms, and a common factor of
3 from the last two terms:
2x(x + 5) +
3(x + 5)
Now, (x + 5) is a common factor. Remove it, and you have
(x + 5)(2x + 3)
This method can be used on all of your remaining problems; DO pay attention to the signs of the numbers.
