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Old 07-26-2007   #1
mesrobyet
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Default Algebra Problem?

(2x^2+5x-3)(4x-1)=ax^2+bx^2+cx+dWhat is the value of b?Please also state how you solved it.SORRY, TYPO! IGNORE ABOVE EQUATION!(2x^2+5x-3)(4x-1)=ax^3+bx^2+cx+d ^ Third, not second.
 
Old 07-26-2007   #2
Paris
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Assuming you could do it without the exponents...you have to match them up according to that exponent.5^0 is different than 5^1
 
Old 07-26-2007   #3
diburning
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It's hard to explain, but here's the answer8x^3)+(18x^2)-12x+3Since the form is (Ax^3)+(Bx^2)+Cx+DB = 18x^2
 
Old 07-26-2007   #4
ttly^tink
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sorry i couldnt shorten it further but i shortened the problem to 8x^3+18x^2-17x+3=Ax^3+Bx^2=Cxwe never did that in my algebra class this year 0.0 but i did the most i cud srrrrrrry
 
Old 07-26-2007   #5
Kevin B
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You could go through the trouble of multiplying out your terms, or just isolate on what will give you x^2 terms - in this case, 4x * 5x, and 2x^2 * (-1), which gives you 20x^2-2x^2, or 18x^2. Therefore, b = 18.
 
 

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