Multiplying (2y - 4)(y - 3)(4y + 5)

[Ladybug]

(2y-4)(y-3)(4y+5)

How do I go about multiplying this? I have five answer choices and I'm getting none of them. I did it by using the FOIL method for two and then mulitiplying vertically for the results, but the numbers come out huge, like 200y and 240y. So I know its wrong, b/c the choices I have are like 8y cubed and 50 y squared.

What am I doing wrong?

Thanks very much.

 

[skeeter]

I have no idea what you are "doing wrong" because you have not shown how you expanded the product.

(2y - 4)(y - 3)(4y + 5) =

(2y² - 10y + 12)(4y + 5) =

8y³ - 30y² - 2y + 60

 

[Ladybug]

I have no idea what you are "doing wrong" because you have not shown how you expanded the product.

(2y - 4)(y - 3)(4y + 5) =

(2y² - 10y + 12)(4y + 5) =

8y³ - 30y² - 2y + 60

Oh, sorry about that.

I think all I need is an explanation of how to simplify (2y² - 10y + 12)(4y + 5). That's what I'm having trouble on. Is it 4y(2y² - 10y + 12) and then 5(2y² - 10y + 12)? That's how I'm getting such huge numbers. Please tell me what I need to do here. I know I'm doing something quite wrong. Thanks!

 

[stapel]

I think all I need is an explanation of how to simplify (2y² - 10y + 12)(4y + 5). That's what I'm having trouble on.

I'll bet good money you were taught "FOIL", as though that were the general method for multiplying polynomials, and that's why you're lost. I truly wish they'd dump this extremely-misleading "short-cut"! :?

To learn how to multiply polynomials in general, try studying some of the many great lessons available online. They all have worked examples, which will go a long way toward explaining the methods and steps. If you remember how to do long (vertical) numerical multiplication, you'll figure this out in no time! :wink:

Eliz.

 

[jwpaine]

FOIL two of the binomial factors into an expansion and then multiply the remaining binomial by all previously expanded terms resulting in a final expansion.

John

 

[Mrspi]

I think all I need is an explanation of how to simplify (2y² - 10y + 12)(4y + 5). That's what I'm having trouble on. Is it 4y(2y² - 10y + 12) and then 5(2y² - 10y + 12)? That's how I'm getting such huge numbers. Please tell me what I need to do here. I know I'm doing something quite wrong. Thanks!

You've got it right....the product of the first two binomials is (2y<SUP>2</SUP> - 10y + 12).

Now, you need to multiply (2y<SUP>2</SUP> - 10y + 12) times (4y + 5)

Multiply each term in the first set of parentheses by each term in the second set of parentheses:

(2y<SUP>2</SUP> - 10y + 12) * (4y + 5)

(2y<SUP>2</SUP> - 10y + 12) * 4y + (2<SUP>2</SUP> - 10y + 12)*5

2y<SUP>2</SUP>*4y - 10y*(4y) + 12*4y + 2y<SUP>2</SUP>*5 - 10y * 5 + 12 * 5

Do the multiplications, and combine like terms.....