factoring completly
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ALGEBRA
3x2 = 17x + 6
asked by Randy on April 16, 2012 
Math
1.3x^2+17x+10 2.5y^8125 3.a^22ab15b^2 *Please help me.You have to factor by completing. Let's look at #1: 3x^2 + 17x + 10 We have 2 and 5 as factors of the last term. What combinations can we do to arrive at the middle term?
asked by Alex on February 4, 2007 
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The parabolas y = 3x2 + 10x  6 and y = 3x2  17x + 2 intersect at _____? A.the line y = 27x  8. B. x=the square root of 2 over 3 and xthe square root of 2 over 3. C. x=8/27. D. they do not intersect.
asked by Tifini on July 8, 2012 
college math
use long division 1.2x3+3x217x+1 /x2 2.y3+3y2+1 /y2 3.n5n2 /n2n
asked by erin on November 11, 2011 
Algebra
(x)/(x3)4(2x5)/(x+2) I posted something before this, so now this is how far I got. The answer I know is (5x^217x9)/(x3)(x+2) first I get (x)(x+2)/(x3)(x+2) (4)(x3)(x+2)/(x3)(x+2)(2x5)(x3)/(x3)(x+2) Then I get
asked by Sara on July 20, 2007 
Algebra
(x)/(x3)4(2x5)/(x+2) I posted something before this, so now this is how far I got. The answer I know is (5x^217x9)/(x3)(x+2) first I get (x)(x+2)/(x3)(x+2) (4)(x3)(x+2)/(x3)(x+2)(2x5)(x3)/(x3)(x+2) Then I get
asked by Sara on July 20, 2007 
statistics
Use the list and calculation method to answer the following questions. Show your work at each step. 4. You draw three cards (with replacement) from a standard deck of cards. What is the probability that a. exactly one will be red:
asked by Lynn on June 7, 2013 
algebra
Can someone please explain What polynomial has a graph that passes through the given points? (–4, 89), (–3, 7), (–1, –1), (1, –1), (4, 329) y = 2x3 – 3x2 – 2x + 1 y = 1x4 – 2x3 – 3x2 + 2x + 1 y = x4 – 2x3 + 3x2
asked by lee on November 14, 2012 
Math help please check
Add or Subtract for #34 3. (2x2 + 6x + 1) + (–7x2 + 2x – 3) 5x2 – 4x – 2 –5x2 + 8x – 2 5x2 – 8x + 2 –9x2 – 8x + 2 4. (3x2 – 7x – 4) – (6x2 – 6x + 1) –3x2 – x –5 –3x2 –13x + 5 9x2 – x + 5 3x2
asked by Delilah on May 6, 2013 
Algebra
Solve: 5x3=11+12x; 17x3=11; 17x3+3=11+3= 17x=14; x= 14/17 is this right?
asked by A.W. on February 1, 2009