One of the roots of 2x 3 − 9x 2 + 13x + k = 0 is x = 2. Find the other two roots.
+505 First, to find out what k is, just use the remainder theorem: if x=2 is a root, then when you evaluate the polynomial at 2 you should get zero. Plugging 2 in for x:
\(2\cdot2^3 - 9\cdot2^2+13\cdot2+k = 0\\6+k=0\\k=-6\)
Knowing that, just use polynomial division to divide this into a quadratic:
\(\hspace{2.35cm}2x^2 - 5x + 3 \\x-2|\overline{2x^3-9x^2+13x-6}\\ \hspace{1.15cm}2x^3 - 4x^2\\ \hspace{2cm} \overline{-5x^2+13x}\\ \hspace{1.8cm} -5x^2 + 10x\\ \hspace{3.75cm} \overline{3x-6}\\ \hspace{3.75cm}3x-6 \\ \hspace{4.75cm} \overline{0}\)
Then, factor the quadratic:
\(2x^2-5x+3 = (2x-3)(x-1)\)
Therefore, the two other roots are \(\boxed{\frac{3}{2}, 1}\)
