I need help with a math question
- Thread starter RA1D3R
- 25 comments
- 32,647 views
- 8,343
- Auckland
- Lazer0pz
Basic Quadratics...
So the equation is Y = 3[(x+5)(x-2)].
First, multiply (x+5)(x-2) to get x²+3x-10
Y = 3[(x+5)(x-2)]
∴ Y = 3[x²+3x-10]
Multiply everything in the bracket by what's outside the bracket..
∴ Y = 3x²+9x-30
Waiting for someone to realise that they've been tree'd
So the equation is Y = 3[(x+5)(x-2)].
First, multiply (x+5)(x-2) to get x²+3x-10
Y = 3[(x+5)(x-2)]
∴ Y = 3[x²+3x-10]
Multiply everything in the bracket by what's outside the bracket..
∴ Y = 3x²+9x-30
Waiting for someone to realise that they've been tree'd
- 7,527
- Exorcet
- OE Exorcet
Order of operations is PEMDAS
Parenthesis
Exponents
Multiply
Divide
Add
Subtract
P
y = 3 ( x + 5 ) ( x - 2 )
since x is an unknown, we can't do anymore in the ()'s
E
No exponents
M
y = 3 ( x + 5 ) ( x - 2 ) = (3x + 15)(x - 2) = 3x^2 + 15x - 6x - 30
D
None
A/S
3x^2 + 15x - 6x - 30 = 3x^2 + 9x - 30
If I pretend not to see the post above me am I tree'd?
Parenthesis
Exponents
Multiply
Divide
Add
Subtract
P
y = 3 ( x + 5 ) ( x - 2 )
since x is an unknown, we can't do anymore in the ()'s
E
No exponents
M
y = 3 ( x + 5 ) ( x - 2 ) = (3x + 15)(x - 2) = 3x^2 + 15x - 6x - 30
D
None
A/S
3x^2 + 15x - 6x - 30 = 3x^2 + 9x - 30
If I pretend not to see the post above me am I tree'd?
- 28,419
- Glasgow
- GTP_Mars
The rules of multiplying out quadratics is FOIL (first - outer - inner - last).. and multiply out the brackets first...
With (x + 5) (x - 2):
Your 'firsts' are x * x = x²
Your 'outers' are x * -2 = -2x
Your 'inners' are 5 * x = (+)5x
Your 'lasts' are 5 * -2 = -10
So, (x + 5) (x - 2) becomes: [x² - 2x + 5x - 10] which equals x² + 3x - 10
Multiply the whole lot by 3, and you get:
y = 3x² + 9x - 30
--
@ Exorcet - not as tree'd as me
- 1
- United States
why do you multiply the whole thing by 3?The rules of multiplying out quadratics is FOIL (first - outer - inner - last).. and multiply out the brackets first...
With (x + 5) (x - 2):
Your 'firsts' are x * x = x²
Your 'outers' are x * -2 = -2x
Your 'inners' are 5 * x = (+)5x
Your 'lasts' are 5 * -2 = -10
So, (x + 5) (x - 2) becomes: [x² - 2x + 5x - 10] which equals x² + 3x - 10
Multiply the whole lot by 3, and you get:
y = 3x² + 9x - 30
--
@ Exorcet - not as tree'd as me
- 84,445
- Rule 12
- GTP_Famine
The original equation is y = 3 (x+5)(x-2)
@Touring Mars solved (x+5)(x-2). Multiplying it all by 3 then solves 3(x+5)(x-2).
- 2,226
- Bács-Kiskun
The Holy Mother of necroposts is among us!
- 34,813
- NoDak
- Cy-Fi
And they signed up today to ask that...
GTV0819
(Banned)
- 6,084
- Papua New Guinea
Oh please.
Sorry, huh? I forgot to put there that it was more than 7 years ago since I've been carried away for looking at the actual date of the original post but at least I do know that it isn't exactly 8 years yet.
Come on, just because I simply said '7' and not '8', you already became so merely amused about it.
GTV0819
(Banned)
- 6,084
- Papua New Guinea
Danoff
Premium
- 33,003
- Mile High City
Since you've been curious about this for so many years, I figured it's worth explaining a little further. Alternatively you could have multiplied just the first one set of brackets by 3. For example:
(3x+15) (x -2)
or another alternative
(x+5) (3x-6)
Edit:
For completeness:
also (x -2)(3x+15)
also (3x-6)(x+5)
Doesn't matter what order you perform the multiplication in.
- 5,014
- Panama City, FL
My algebra teacher tought us to multiply two binomials with a "monkey-face." 
OK, this was a L-O-N-G time ago...
Write the expression down:
Then draw lines above from the first and second term in the first set to the first and second term in the second set. then draw lines below between the inner terms and the outer terms.
Do the multiplication for each line you drew and add them all together.
As to why you multiply the whole thing by 3, well, it's because "x-squared plus 3x minus 10" is ONE number, not 3. You can't multiply by part of a term.
OK, this was a L-O-N-G time ago...Write the expression down:
Then draw lines above from the first and second term in the first set to the first and second term in the second set. then draw lines below between the inner terms and the outer terms.
Do the multiplication for each line you drew and add them all together.
As to why you multiply the whole thing by 3, well, it's because "x-squared plus 3x minus 10" is ONE number, not 3. You can't multiply by part of a term.
Danoff
Premium
- 33,003
- Mile High City
My algebra teacher tought us to multiply two binomials with a "monkey-face."
Write the expression down:
Then draw lines above from the first and second term in the first set to the first and second term in the second set. then draw lines below between the inner terms and the outer terms.
Do the multiplication for each line you drew and add them all together.
As to why you multiply the whole thing by 3, well, it's because "x-squared plus 3x minus 10" is ONE number, not 3. You can't multiply by part of a term.
I feel like this method and the foil method are actually somewhat counter-productive and difficult to remember. Doesn't it make more sense to just distribute the multiplication?
(x+5)(x-2)
becomes
x(x-2)+5(x-2)
x^2-2x+5x-10 => x^2+3x-10
So 3(x+5)(x-2) can be distributed as (3x+15)(x-2) which can be distributed as 3x(x-2)+15(x-2) which is (3x^2+9x-30).
I don't know if it's true for anyone else, but this makes a lot more sense to me than trying to remember to do foil or anything else, and it even scales better to harder problems.
