Solve 2k³+15k²+37k+24/24= [(k+1)(2(k+1)²+9(k+1)+13)/24]

Question

Please solve the equation

Answer 1

2k³+15k²+37k+24/24= [(k+1)(2(k+1)²+9(k+1)+13)/24]

Expand the right-hand side: (2(k+1)^3+9(k+1)^2+13(k+1))/24.

The denominators are the same on each side so they can be cancelled out.

2((k+1)^3-k^3)+12(k+1)^2-12k^2-3(k+1)^2-3k^2+13k+13-37k-24=0;

2(3k^2+3k+1)+12(2k+1)-3(2k^2+2k+1)-24k-11=0;

6k^2+6k+2+24k+12-6k^2-6k-3-24k-11=0;

0=0; so the problem was in fact an identity, both sides being identically equivalent.

CHECK

Put k=0: 24=[2+9+13]. OK.

Put k=-1: -2+15-37+24=0. OK.

Answer 2

2k³ + 15k ² + 37k + 24/24 = [(k+1)(2(k+1)²+9(k+1)+13)/24]

=> 23k^3/12 + 115k^2/8 + 815k/24 = 0

=> k = 0

k = ±1/4 i (√71±15i)

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Comments

I think you should find that the left and right sides of this are identical, therefore true for all k. The easiest way to see this is to substitute for k on each side. Choose easy values for k, like 0, 1, -1. You only need a couple of values to prove the identity. For example, k=1:

78/24=2(8+18+13)/24=78/24. With k=0 and -1 it's even easier!

Check your calculations, because there must be an error somewhere.