Q1: Use synthetic division method to find all the zeros of the polynomial: f(x) = x4 + x3 - 11x2 - 5x + 30 Q2: Employ the Newton-Raphson method and determine the zeros of the polynomial: f(x) = 0.5x4 − 4x3 + 6x − 2 using initial guess of 4.2 to a level of accuracy of +-10⁻¹

According to the Rational Root Theorem, any rational root $\frac{p}{q}$ must have $p$ as a divisor of the constant term (30) and $q$ as a divisor of the leading coefficient (1).
Divisors of 30: $\pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30$.
Since the leading coefficient is 1, the possible rational roots are these divisors.

Let's test $x=2$:

Since the remainder is 0, $x=2$ is a zero of the polynomial. The depressed polynomial is $x^3 + 3x^2 - 5x - 15$.

We can factor $x^3 + 3x^2 - 5x - 15$ by grouping:

Set the factors to zero to find the remaining roots:

The zeros of the polynomial are $2, -3, \sqrt{5},$ and $-\sqrt{5}$.

The zeros of the polynomial are $\boxed{2, -3, \sqrt{5}, -\sqrt{5}}$.

Q2: Employ the Newton-Raphson method and determine the zeros of the polynomial: $f(x) = 0.5x^3 - 4x^2 + 6x - 2$ using initial guess of 4.2 to a level of accuracy of $\pm 10^{-1}$.

Given $f(x) = 0.5x^3 - 4x^2 + 6x - 2$.
The derivative is $f'(x) = \frac{d}{dx}(0.5x^3 - 4x^2 + 6x - 2) = 1.5x^2 - 8x + 6$.

The initial guess is $x_0 = 4.2$. The accuracy required is $|x_{n+1} - x_n| < 0.1$.

• Iteration 1: $x_0 = 4.2$ $f(x_0) = f(4.2) = 0.5(4.2)^3 - 4(4.2)^2 + 6(4.2) - 2 = 37.044 - 70.56 + 25.2 - 2 = -10.316$ $f'(x_0) = f'(4.2) = 1.5(4.2)^2 - 8(4.2) + 6 = 26.46 - 33.6 + 6 = -1.14$ $x_1 = 4.2 - \frac{-10.316}{-1.14} = 4.2 - 9.04912 = -4.84912$ $|x_1 - x_0| = |-4.84912 - 4.2| = 9.04912 > 0.1$

• Iteration 2: $x_1 = -4.84912$ $f(x_1) = 0.5(-4.84912)^3 - 4(-4.84912)^2 + 6(-4.84912) - 2 = -56.9895 - 94.0552 - 29.0947 - 2 = -182.1394$ $f'(x_1) = 1.5(-4.84912)^2 - 8(-4.84912) + 6 = 35.2707 + 38.7929 + 6 = 80.0636$ $x_2 = -4.84912 - \frac{-182.1394}{80.0636} = -4.84912 - (-2.27494) = -2.57418$ $|x_2 - x_1| = |-2.57418 - (-4.84912)| = 2.27494 > 0.1$

• Iteration 3: $x_2 = -2.57418$ $f(x_2) = 0.5(-2.57418)^3 - 4(-2.57418)^2 + 6(-2.57418) - 2 = -8.5204 - 26.5041 - 15.4451 - 2 = -52.4696$ $f'(x_2) = 1.5(-2.57418)^2 - 8(-2.57418) + 6 = 9.9392 + 20.5934 + 6 = 36.5326$ $x_3 = -2.57418 - \frac{-52.4696}{36.5326} = -2.57418 - (-1.43624) = -1.13794$ $|x_3 - x_2| = |-1.13794 - (-2.57418)| = 1.43624 > 0.1$

• Iteration 4: $x_3 = -1.13794$ $f(x_3) = 0.5(-1.13794)^3 - 4(-1.13794)^2 + 6(-1.13794) - 2 = -0.7370 - 5.1799 - 6.8276 - 2 = -14.7445$ $f'(x_3) = 1.5(-1.13794)^2 - 8(-1.13794) + 6 = 1.9425 + 9.1035 + 6 = 17.0460$ $x_4 = -1.13794 - \frac{-14.7445}{17.0460} = -1.13794 - (-0.86498) = -0.27296$ $|x_4 - x_3| = |-0.27296 - (-1.13794)| = 0.86498 > 0.1$

• Iteration 5: $x_4 = -0.27296$ $f(x_4) = 0.5(-0.27296)^3 - 4(-0.27296)^2 + 6(-0.27296) - 2 = -0.0102 - 0.2980 - 1.6378 - 2 = -3.9460$ $f'(x_4) = 1.5(-0.27296)^2 - 8(-0.27296) + 6 = 0.1118 + 2.1837 + 6 = 8.2955$ $x_5 = -0.27296 - \frac{-3.9460}{8.2955} = -0.27296 - (-0.47566) = 0.20270$ $|x_5 - x_4| = |0.20270 - (-0.27296)| = 0.47566 > 0.1$

• Iteration 6: $x_5 = 0.20270$ $f(x_5) = 0.5(0.20270)^3 - 4(0.20270)^2 + 6(0.20270) - 2 = 0.0042 - 0.1643 - 1.2162 - 2 = -0.9440$ $f'(x_5) = 1.5(0.20270)^2 - 8(0.20270) + 6 = 0.0616 - 1.6216 + 6 = 4.4400$ $x_6 = 0.20270 - \frac{-0.9440}{4.4400} = 0.20270 - (-0.21261) = 0.41531$ $|x_6 - x_5| = |0.41531 - 0.20270| = 0.21261 > 0.1$

• Iteration 7: $x_6 = 0.41531$ $f(x_6) = 0.5(0.41531)^3 - 4(0.41531)^2 + 6(0.41531) - 2 = 0.0359 - 0.6906 + 2.4919 - 2 = -0.1628$ $f'(x_6) = 1.5(0.41531)^2 - 8(0.41531) + 6 = 0.2588 - 3.3225 + 6 = 2.9363$ $x_7 = 0.41531 - \frac{-0.1628}{2.9363} = 0.41531 - (-0.05544) = 0.47075$ $|x_7 - x_6| = |0.47075 - 0.41531| = 0.05544 < 0.1$

Since $|x_7 - x_6| < 0.1$, we take $x_7$ as the approximation.
Rounding 0.47075 to one decimal place (accuracy of $\pm 10^{-1}$), we get 0.5.

The zero of the polynomial using the Newton-Raphson method with an initial guess of 4.2 to an accuracy of $\pm 10^{-1}$ is $\boxed{0.5}$.