9x 2+6x+1 = (3x+1)(3x+1)=(3x+1) 2 (Hint: You might need to try more than one pair of factors for 9x.
2x 2+3x-5 = (2x+5)(x-1) ← highlight to see answer solving quadratic, answers to kuta software infinite algebra 1, free algebra 2 worksheets kuta software llc, kuta software infinite algebra 2 logarithmic equations 1, infinite algebra 2 assignment.These problems can't be factored into easier equations, so you'll need to work out an answer in the form of (_x + _)(_x + _) by trial and error: 3x 3+3x 2-6x = (3x)(x+2)(x-1) ← highlight that space to see the answer.Highlight the space after the equal signs to see the answer so you can check your work: These problems have a common factor in each term that needs to be factored out first. These are the problems from the step about "trickier factors." We already simplified them down to an easier problem, so try to solve them using the steps in method 1, then check your work here: Answers to "tricky factoring" problems.This won't work for more complicated problems, though, so it's good to remember the "long way" described above. In simple cases like this, when you don't have a constant in front of the x 2 term, you can use a shortcut: just add the two factors together and put an "x" after it (-2+5 → 3x).That matches the original polynomial, so this is the correct answer: (x-2)(x+5). In fact, once you test -1 and 10, you know that 1 and -10 will just be the opposite of the answer above: -9x instead of 9x. Our original problem has an "x" term of 3x, so that's what we want to end up with in this test.Use trial and error to test each possibility, multiplying the Outside and Inside terms, and comparing the result to our trinomial. We've narrowed the Last terms down to a few possibilities. Test which possibilities work with Outside and Inside multiplication.