Factor the polynomial completely. If it cannot be factored, write Prime.?

Prime...........................Show more

= 3k² - 2k - 7

= 3.[k² - (2/3).k - (7/3)]

= 3.[k² - (2/3).k + {(1/3) - (1/3)} - (7/3)]

= 3.[k² - (2/3).k + (1/3)² - (1/3)² - (7/3)]

= 3.[{k² - (2/3).k + (1/3)²} - (22/9)]

= 3.[{k - (1/3)}² - (22/9)]

= 3.[{k - (1/3)}² - {(√22)/3}²] → recall: a² - b² = (a + b).(a - b)

= 3.[{k - (1/3)} + {(√22)/3}].[{k - (1/3)} - {(√22)/3}]

= 3.[k - (1/3) + (√22)/3].[k - (1/3) - (√22)/3]

= 3.[k - (1 - √22)/3].[k - (1 + √22)/3] → to go further, if you want it:

= [3k - (1 - √22)].[k - (1 + √22)/3]

= [3k - 1 + √22)].[k - (1 + √22)/3]...Show more

3k^2 - 2k - 7
= 1/3 (3 k - 1)^2 - 22/3...Show more

That polynomial has irrational roots so cannot be factored over rational numbers.  It's prime.

A test is to check the discriminant and see if it's a negative, positive perfect square, or positive and not a square.

If negative, no real roots.  Can't be factored.
If positive but not a square, can't be factored.
If positive but a square, it can be factored.

b² - 4ac
(-2)² - 4(3)(-7)
4 + 84
88

This is not a perfect square so cannot be factored....Show more