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 March 12th, 2009, 07:36 PM #1 Newbie   Joined: Mar 2009 Posts: 24 Thanks: 0 homework help! how do i times these? i know i use foil but fractions always confuse me! (2x^2-5x+(x/3-8/3)
 March 12th, 2009, 08:05 PM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: homework help! $a \times \frac{c}{d}= \frac{ac}{d}$ So... $\\(2x^2-5x+(\frac13x-\frac83) \\=(2x^2-5x+\frac13x - (2x^2-5x+\frac83 \\=\frac23x^3-\frac53x^2+\frac83x \quad - \quad \frac{16}{3}x^2 + \frac{40}{3}x-\frac{64}{3} \\\mbox{collecting like terms...} \\=\frac23x^3-\frac{21}{3}x^2+\frac{48}{3}x-\frac{64}{3} \\=\frac23x^3-7x^2+16x-21\frac13" />
 March 12th, 2009, 08:40 PM #3 Newbie   Joined: Mar 2009 Posts: 24 Thanks: 0 Re: homework help! ok but home 2x^2 became 2/3x^3?
 March 12th, 2009, 08:48 PM #4 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: homework help! $\\(2x^2-5x+\frac13x \\=\frac23x^3-\frac53x^2+\frac83x" /> The first term in the brackets: $2x^2$ Multiplied by $\frac13x$ $\\=2\cdot\frac13\cdot x \cdot x^2 \\=\frac23 x^3$
March 13th, 2009, 05:12 AM   #5
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Quote:
 Originally Posted by andi7 how do i times these? i know i use foil but fractions always confuse me!
Actually, "FOIL" works only for the one special case in which you're multiplying two binomials. Since that's not what you're doing here, FOIL would not apply!

Quote:
 Originally Posted by andi7 (2x^2-5x+(x/3-8/3)
A better method for this multiplication might be to use vertical multiplication. Set up the factors vertically:

Code:
                2x^2 -      5x +    8
(1/3)x -  8/3
-------------------------------------
Multiply the -8/3 on the top row:

Code:
                2x^2 -      5x +    8
(1/3)x -  8/3
-------------------------------------
- (16/3)x^2 + (40/3)x - 64/3
Multiply the (1/3)x on the top row:

Code:
                2x^2 -      5x +    8
(1/3)x -  8/3
-------------------------------------
- (16/3)x^2 + (40/3)x - 64/3
(3/2)x^3 -  (5/3)x^2 +  (8/3)x
-------------------------------------

 March 13th, 2009, 07:55 AM #6 Global Moderator   Joined: Dec 2006 Posts: 21,119 Thanks: 2331 (2x² - 5x + 8)(x/3 - 8/3) = (2x² - 5x + 8)(x - 8)/3 (this defers working with fractions). Each term within the second set of parentheses has to be applied to each term within the first set (or vice versa). We systematically choose the pairs of terms that will combine to produce as high a power of x as possible (without choosing the same pair again, of course), and combine like powers. The results are written within parentheses and then the division by 3 is appended. Thus, one writes (2x² * x + 2x² * (-8) - (5x)x + 8*x - (5x) * (-8) + 48x + 8 * (-8))/3, which simplifies to (2x³ - 21x² + 48x - 64)/3, which equals (2/3)x³ - 7x² + 16x - 64/3. The final term can be written as 21 1/3 if you like. With practice, much of the above can be done mentally, allowing the entire calculation to be written on one line.
March 13th, 2009, 11:55 AM   #7
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Re: homework help!

Quote:
 Originally Posted by stapel A better method for this multiplication might be to use vertical multiplication. Set up the factors vertically
I've never seen it done that way, that's kinda neat. Just goes to show, a million ways to do the same thing.

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