Question

Verify whether 2, 3 and $\frac{1}{2}$ are the zeroes of the polynomial 𝒑(𝒙)= $2x^3–11x^2+17x–6$.

Answer 1

Step 1:
Substitute x = 2 in p(x).
P(2) = $2(2{)}^3-11(2{)}^2+17\left(2\right)-6$
P(2) = 16 - 44 + 34 - 6
P(2) = 0

Hence, it is verified that x = 2 is a zero, since on substituting the value of x = 2 in p(x), we got p(x) = 0.

Step 2:
Substitute x = 3 in p(x).
Substitute x = 2 in p(x).

P(3) = $2(3{)}^3-11(3{)}^2+17\left(3\right)-6$
P(3) = 54 - 99 + 51 - 6
P(3) = 0

Hence, it is verified that x = 3 is a zero, since on substituting the value of x = 3 in p(x), we got p(x) = 0.

Step 3:
Substitute x = $\frac{1}{2}$ in p(x).


Hence, it is verified that x = 1/2 is a zero, since on substituting the value of x = 1/2 in p(x), we got p(x) = 0.

Hence we say that 2, 3 and ½ are the zeroes of the given polynomial.