Problem 33 - Beginning and Intermediate Algebra

Question

Factor by trial and error. $$3 u^{2}-23 u+30$$

Step-by-Step Solution

Answer

The factored expression for \(3u^2 - 23u + 30\) is \((3u+5)(u+6)\).

Step 1: Identify possible factors for the first and last terms

For the first term, \(3u^2\), the possible factorizations are: 1. \(u\) and \(3u\) For the last term, \(30\), the possible factorizations are: 1. \(1\) and \(30\) 2. \(-1\) and \(-30\) 3. \(2\) and \(15\) 4. \(-2\) and \(-15\) 5. \(3\) and \(10\) 6. \(-3\) and \(-10\) 7. \(5\) and \(6\) 8. \(-5\) and \(-6\)

Step 2: Test possible factor combinations for the correct middle term

Now, we'll create pairs of the possible factors, and test them to see if they produce the correct middle term. 1. \((u+1)(3u+30)\) produces a middle term of \(33u\) 2. \((u-1)(3u-30)\) produces a middle term of \(-27u\) 3. \((u+2)(3u+15)\) produces a middle term of \(21u\) 4. \((u-2)(3u-15)\) produces a middle term of \(-9u\) 5. \((u+3)(3u+10)\) produces a middle term of \(39u\) 6. \((u-3)(3u-10)\) produces a middle term of \(-3u\) 7. \((u+5)(3u+6)\) produces a middle term of \(-3u + 15u = 12u\) 8. \((u-5)(3u-6)\) produces a middle term of \(-15u + 3u = -12u\) 9. \((3u+1)(u+30)\) produces a middle term of \(93u\) 10. \((3u-1)(u-30)\) produces a middle term of \(-87u\) 11. \((3u+2)(u+15)\) produces a middle term of \(47u\) 12. \((3u-2)(u-15)\) produces a middle term of \(-33u\) 13. \((3u+3)(u+10)\) produces a middle term of \(63u\) 14. \((3u-3)(u-10)\) produces a middle term of \(-21u\) 15. \((3u+5)(u+6)\) produces a middle term of \(23u\) 16. \((3u-5)(u-6)\) produces a middle term of \(-18u\) So we can see that the correct factored form is in option 15.

Step 3: Write the correct factored form

The correct factored form is \((3u+5)(u+6)\). So, the factored expression for \(3u^2 - 23u + 30\) is \((3u+5)(u+6)\).