f(x)=5x^5−50x ^4+63x^3+330x ^2+36x+216, After doing all layers of synthetic division from the known rational roots, what factor remains? Hint* There is a double root at 6

Question

i need help icant figure out how to do this

Answer 1

Give two roots at 6, so we can start synthetic division with 6

5 -50 63 330 36 216 |6

30 -120 -342 -72 -216

—————————————

5 -20 -57 -12 -36 0

5 -20 -57 -12 -36 |6

30 60 18 36

——————————

5 10 3 6 0

This leaves 6 as the last coefficient. Factors of 6 are ±1,2,3,6

Try -2

5 10 3 6 |-2

-10 0 -6

——————

5 0 3

This leaves 5x²+3 --> x = i√3/5, -i√3/5

f(x) = (x-6)(x-6)(x+2)(5x²+3)

Answer 2

Well, the hint helps a lot. Also I would immediately graph the equation to get an idea about where the roots lie.

Use long division or synthetic division to divide out the double root (2 divisions by x-6).

This will leave you with a 4th degree polynomial with a constant term of 6.

Use the rational root theorem to look for other roots: the numerator of any rational root must divide 6...so try ±1, ±2, ±3, and ±6.

Also the denominator of any rational root must divide 5.

This should get you some more roots OR exclude rational roots.

If you find rational roots, divide them out by long division.

If you are left with a quadratic, you should know what to do.

If you are left with either a cubic or a quartic, the graph should tell you how many real roots there are and you can use Descartes rule of signs to help ascertain that.