x^3 – 5x +3x -15 – Factoring by group

In order to factor by grouping this polynomial we have to see if the terms have any GCF (Greatest Common Factor). In this case there is 1 GCF. Then we create smaller groups by grouping the first two terms together and the last two terms together which will be x^3 – 5x^2 + 3x – 15, then we factor out the GCF from each of the two groups and in this problem the signs in front of the 5x^2 and the 15 are different so you need to factor out a positive 3 which then you will get x^2(x – 5) + 3(x – 5), the thing that the groups have in common are the (x – 5) so you can factor that out which gives you (x – 5)(x^2+3), then see if any of the remaining factors can be factored which they cannot so that is the final answer

18x^2 – 98y^2 – Factoring the difference of squares

the two terms have 2 in common and gives us 2(9x^2 – 49^2), To factor this problem into the form (a + b)(a – b), you need to determine what squares will equal 9x^{2}and what squared will equal 49y^{2}. In this case the choices are 3x and 7y because (3x)(3x) = 9x^{2} and (7y)(7y) =49y^2 which gives us 2(3x + 7y)(3x – 7y)

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Tags: factoring, Math, precalc, week4-B