Factoring help

[betta14]

1.How do you factor (-5t^2+50t+5) ? I keep getting (-5)(t-1)(t+1), but when I check it by foiling and multiplying, it doesn't equal the original trinomial. 2. Can I multiply (-3x-2x+8) by (-1) to get rid of the first negative coefficient? If I don't, it will be differently factored, but will still be "correct". Which one is the right answer?

 

[mathwannabe]

1.How do you factor (-5t^2+50t+5) ? I keep getting (-5)(t-1)(t+1), but when I check it by foiling and multiplying, it doesn't equal the original trinomial. 2. Can I multiply (-3x-2x+8) by (-1) to get rid of the first negative coefficient? If I don't, it will be differently factored, but will still be "correct". Which one is the right answer?

1) Here is the problem, rewritten:

\(\displaystyle (-5t^2+50t+5)\)

To factor it you can use the distributive law which says: \(\displaystyle ab + ac = a(b+c)\) ,translated to simple "real-world" examples:

\(\displaystyle 2(7)+2(9)=2(7+9)=32\)

\(\displaystyle 6+14=2(3+7)=20\)

\(\displaystyle 9+18=3(3+6)=9(1+2)=27\)

So, your problem:

\(\displaystyle (-5t^2+50t+5)=5(-t^2+10t+1)=-5(t^2-10t-1)\)

2) Why?

\(\displaystyle (-3x-2x+8)=-5x+8\)

 

[lookagain]

2. Can I multiply (-3x-2x+8) by (-1) to get rid of the first negative coefficient? If I don't, it will be differently factored, but will still be "correct". Which one is the right answer?

betta14,

didn't you intend to type \(\displaystyle " -3x^2 - 2x + 8 \ ?"\)


That would be consistent with the trinomial you asked about in the first question.