Because we are in the chapter of Factoring Polynomials and in the section called"Using Several Methods of Factoring" and I don't know how to do it.
Another problem I don't get is: -41a+10+21a^2
2007-01-07 07:41:25 · 8 answers · asked by snape_fan_2005 2 in Science & Mathematics ➔ Mathematics
5a^2+10ab+5b^2
5(a^2+2ab+b^2)
5(a+b)^2
21a^2-41a+10
(3a-5)(7a-2)
2007-01-07 07:47:24 · answer #1 · answered by 7 · 0⤊ 0⤋
Look for common numbers and variables.
5a^2+10ab+5b^2
5(a^2+2ab+b^2)
Then factor the tri nomial.
2007-01-07 15:44:13 · answer #2 · answered by Anonymous · 0⤊ 0⤋
The first proble I can't help you with I suck at factoring but the second problem I can help you with. rewrite it like this:
21x^2 - 41x + 10 = 0
use this fromula to solve it:x = -b + or - sqr(b^2 - 4ac)/2a
where a & b are the coefiecents of x^2 and x repsectively and c is an integer
2007-01-07 16:14:41 · answer #3 · answered by ikeman32 6 · 0⤊ 0⤋
add the whole numbers together and then put the vaviables like this
a^2+b^2 and put that at the end of the sum of the whole numbers
so it should be 20a^2+b^2
as for the other one ill just give u the answer
-10a^3
2007-01-07 15:47:32 · answer #4 · answered by Wes 2 · 0⤊ 0⤋
5a^2+10ab+5b^2
you have to combine like terms
5a+10a+b+5b+2/2
5a+10a=15a THEN b+5b=6b
10a+6b^2 and figure that out! and your done i think.
2007-01-07 15:51:54 · answer #5 · answered by rundumctrickaaay 2 · 0⤊ 0⤋
5(a+b)(a+b)
2007-01-07 15:46:52 · answer #6 · answered by Zidane 3 · 0⤊ 0⤋
=5(a^2+2ab+b^2)=5(a+b)^2=>(a+b)=0
2007-01-07 15:47:27 · answer #7 · answered by arootingh 2 · 0⤊ 0⤋
put it on Yahoo Answers and let someone else do it...that is how you do it
2007-01-07 15:43:29 · answer #8 · answered by tellme 4 · 0⤊ 0⤋