Question

which of these is a possible root of the polynomials? (hint: graph)\\(x^3 - 6x^2 - x + 30\\)\
options: 0, -9, -15, 4

Explanation:

Step1: Apply Rational Root Theorem

Possible roots: $\pm1, \pm2, \pm3, \pm5, \pm6, \pm10, \pm15, \pm30, \pm\frac{1}{1}$

Step2: Test $x=4$ in polynomial

$4^3 - 6(4)^2 - 4 + 30 = 64 - 96 - 4 + 30 = -6$

Step3: Test other options

• $x=0$: $0 - 0 - 0 + 30 = 30$
• $x=-9$: $(-9)^3 - 6(-9)^2 - (-9) + 30 = -729 - 486 + 9 + 30 = -1176$
• $x=-15$: $(-15)^3 - 6(-15)^2 - (-15) + 30 = -3375 - 1350 + 15 + 30 = -4680$

Step4: Identify closest root

Among options, $x=4$ gives the smallest non-zero result, and the actual roots of $x^3-6x^2-x+30$ are $x=-2, 3, 5$; 4 is the only option near a true root.

Answer:

D. 4