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Help with Algebra Factoring Strategy problem


#1

In the two problems below I am having trouble with the negatives and positives in my answers. Am I missing a step somewhere because my answers are positive whereas theirs are negative.


#2

Step 3 of problem one should read 2x(x-3)-5(x-3)

Not sure about the 2nd problem, check your 3rd step, should read -3 not +3
For the answer I get n(5n-3)(n+2), not sure if that’s right.


#3

The one think that I’m confused about is how you changed the +5x to a -5? I thought that for the two equations in parenthesis to match you would have to change the signs such as -5x+15 to +5(x-3). Could you please explain?


#4

When you get to 2x^2-6x-5x+15, split the expression in the middle to 2x^2-6x and -5x+15 then factor out a common term from each which would be for the first 2x and the second -5. First works out to 2x(x-3) and second -5(x-3). so the factored expression is (x-3)(2x-5). Remember -5*-3=positive 15. not sure if I’m being real clear, but I hope that helps some.

2x^2-11x+15
2x^2-6x-5x+15 split 2x^2-6x and -5x+15, factor out common terms
2x(x-3)-5(x-3)
(x-3)(2x-5) If you multiply both terms together, you get 2x^2-11x+15


#5
  1. 2x2-6x-5x+15
    Note: There is a minus sign in front of the 5x. When you factor out the –5 from –5x+15, –5x/–5 +15/–5 equals x–3, which leads to –5(x–3)
  2. 2x(x–3)–5(x–3)
  3. (x–3)(2x–5)

Same thing in the second problem. in line 2 when you factor the last two terms you get –3(n+2)
The final answer is n(n+2)(5n–3)
The given “correct answer” is wrong. multiplying it out will show this.
(n-2)(2n+3) = 2n2 – n – 6 which is not the original polynomial.
I suspect the given “correct answer” is a typo.


#6

Thank you so much!!!