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

Question

Asked on July 30, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-07-30T23:27:46.751Z

Here are the solutions to your questions. Q1: Use synthetic division method to find all the zeros of the polynomial: f(x) = x^4 + x^3 - 11x^2 - 5x + 30 Step 1: Identify possible rational roots. According to the Rational Root Theorem, any rational root (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: ± 1, ± 2, ± 3, ± 5, ± 6, ± 10, ± 15, ± 30. Since the leading coefficient is 1, the possible rational roots are these divisors. Step 2: Test possible roots using synthetic division. Let's test x=2: c|ccccc 2 & 1 & 1 & -11 & -5 & 30 \\ & & 2 & 6 & -10 & -30 \\ & 1 & 3 & -5 & -15 & 0 \\ Since the remainder is 0, x=2 is a zero of the polynomial. The depressed polynomial is x^3 + 3x^2 - 5x - 15. Step 3: Factor the depressed polynomial to find the remaining zeros. We can factor x^3 + 3x^2 - 5x - 15 by grouping: x^3 + 3x^2 - 5x - 15 = x^2(x+3) - 5(x+3) = (x^2 - 5)(x+3) Set the factors to zero to find the remaining roots: x+3 = 0 x = -3 x^2 - 5 = 0 x^2 = 5 x = ± sqrt(5) The zeros of the polynomial are 2, -3, sqrt(5), and -sqrt(5). The zeros of the polynomial are 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 ± 10^-1. Step 1: Find the derivative of the function. Given f(x) = 0.5x^3 - 4x^2 + 6x - 2. The derivative is f'(x) = (d)/(dx)(0.5x^3 - 4x^2 + 6x - 2) = 1.5x^2 - 8x + 6. Step 2: Apply the Newton-Raphson formula x_n+1 = x_n - (f(x_n))/(f'(x_n)) iteratively until the desired accuracy is met. 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 - (-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 - (-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 - (-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 - (-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 - (-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 - (-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 - (-0.1628)/(2.9363) = 0.41531 - (-0.05544) = 0.47075 |x_7 - x_6| = |0.47075 - 0.41531| = 0.05544 < 0.1 Step 3: Round the final answer to the specified accuracy. Since |x_7 - x_6| < 0.1, we take x_7 as the approximation. Rounding 0.47075 to one decimal place (accuracy of ± 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 ± 10^-1 is 0.5. Drop the next question! 📸

Accepted Answer · 2026-07-30T23:27:46.751Z

Here are the solutions to your questions. Q1: Use synthetic division method to find all the zeros of the polynomial: f(x) = x^4 + x^3 - 11x^2 - 5x + 30 Step 1: Identify possible rational roots. According to the Rational Root Theorem, any rational root (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: ± 1, ± 2, ± 3, ± 5, ± 6, ± 10, ± 15, ± 30. Since the leading coefficient is 1, the possible rational roots are these divisors. Step 2: Test possible roots using synthetic division. Let's test x=2: c|ccccc 2 & 1 & 1 & -11 & -5 & 30 \\ & & 2 & 6 & -10 & -30 \\ & 1 & 3 & -5 & -15 & 0 \\ Since the remainder is 0, x=2 is a zero of the polynomial. The depressed polynomial is x^3 + 3x^2 - 5x - 15. Step 3: Factor the depressed polynomial to find the remaining zeros. We can factor x^3 + 3x^2 - 5x - 15 by grouping: x^3 + 3x^2 - 5x - 15 = x^2(x+3) - 5(x+3) = (x^2 - 5)(x+3) Set the factors to zero to find the remaining roots: x+3 = 0 x = -3 x^2 - 5 = 0 x^2 = 5 x = ± sqrt(5) The zeros of the polynomial are 2, -3, sqrt(5), and -sqrt(5). The zeros of the polynomial are 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 ± 10^-1. Step 1: Find the derivative of the function. Given f(x) = 0.5x^3 - 4x^2 + 6x - 2. The derivative is f'(x) = (d)/(dx)(0.5x^3 - 4x^2 + 6x - 2) = 1.5x^2 - 8x + 6. Step 2: Apply the Newton-Raphson formula x_n+1 = x_n - (f(x_n))/(f'(x_n)) iteratively until the desired accuracy is met. 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 - (-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 - (-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 - (-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 - (-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 - (-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 - (-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 - (-0.1628)/(2.9363) = 0.41531 - (-0.05544) = 0.47075 |x_7 - x_6| = |0.47075 - 0.41531| = 0.05544 < 0.1 Step 3: Round the final answer to the specified accuracy. Since |x_7 - x_6| < 0.1, we take x_7 as the approximation. Rounding 0.47075 to one decimal place (accuracy of ± 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 ± 10^-1 is 0.5. Drop the next question! 📸

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