Question

Problems \(107-116\), are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Solve: \(4 x^{3}+3 x^{2}-25 x+6=0\)

Short answer

The roots are \(x = 2, \ x = \frac{1}{4}, \ x = -3\).

Step-by-step solution

- Identify the polynomial

The problem provides the polynomial equation: \(4x^3 + 3x^2 - 25x + 6 = 0\).

- Use the Rational Root Theorem

The Rational Root Theorem states that any possible rational root of the polynomial equation is a factor of the constant term (6) divided by a factor of the leading coefficient (4). The possible rational roots are \( \pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{1}{4}, \pm \frac{3}{4} \).

- Test the possible roots

Use synthetic division or direct substitution to test the possible roots. Begin with easy integers, such as 1, -1, 2, -2, 3, and -3.

- Verify a root using substitution

Starting with \(x = 1\): \(4(1)^3 + 3(1)^2 - 25(1) + 6 = 4 + 3 - 25 + 6 = -12\). Thus, 1 is not a root.

- Continue testing roots

Next, try \(x = 2\): \(4(2)^3 + 3(2)^2 - 25(2) + 6 = 32 + 12 - 50 + 6 = 0\). So, \(x = 2\) is a root.

- Perform synthetic division

Use synthetic division to divide \(4x^3 + 3x^2 - 25x + 6\) by \(x - 2\). The quotient polynomial is \(4x^2 + 11x - 3\).

- Factor the quotient polynomial

Next, factor \(4x^2 + 11x - 3 = 0\). This factors to \((4x - 1)(x + 3) = 0\).

- Solve for remaining roots

Set each factor equal to zero to find the remaining roots: \(4x - 1 = 0\) gives \(x = \frac{1}{4}\), and \(x + 3 = 0\) gives \(x = -3\).

Key concepts

Rational Root Theorem

The Rational Root Theorem helps in identifying potential rational roots of a polynomial equation. For a polynomial like \(4x^3 + 3x^2 - 25x + 6 = 0\), the theorem states that any possible rational root is a factor of the constant term (6) divided by a factor of the leading coefficient (4). Hence, we obtain the potential roots as \( \pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{1}{4}, \pm \frac{3}{4} \). This gives us a starting point to determine which of these, if any, satisfy the polynomial equation.

Synthetic Division

To verify potential roots, synthetic division is a practical method. It simplifies the testing process by reducing the polynomial and aiding in finding actual roots. For instance, to see if \(x = 2\) is a root of \(4x^3 + 3x^2 - 25x + 6\), we divide the polynomial by \(x - 2\) using synthetic division:
1. List coefficients: {4, 3, -25, 6}
2. Use 2 as the divisor (root candidate)
3. Perform the synthetic division operation:
- Bring down the first coefficient (4)- Multiply the divisor (2) by this number (4), and place the result (8) under the second coefficient
- Add this to the second coefficient (3), resulting in 11
- Repeat for the remaining coefficients.
Through this operation, we find the quotient polynomial as \(4x^2 + 11x - 3\) when the remainder is zero.
Hence \(x = 2\) is confirmed as a root.

Factoring Polynomials

After identifying the quotient polynomial post-synthetic division, the next step is factoring. For the quotient polynomial \(4x^2 + 11x - 3\), we look for factor pairs of the terms that multiply to give the original polynomial. Here, it factors into \((4x - 1)(x + 3) = 0\). This step is crucial as it breaks down the polynomial into simpler binomials, making it easier to solve for roots. Factoring correctly allows us to identify the zeroes of these binomials, which are the roots of our original polynomial.

Root Finding

To find the roots of the polynomial equation, equate each factor to zero. For our factored quotient \((4x - 1)(x + 3) = 0\):

• Setting \(4x - 1 = 0\) leads to \(x = \frac{1}{4}\)
• Setting \(x + 3 = 0\) gives \(x = -3\)

Hence, the roots of \(4x^2 + 11x - 3 = 0\) are \( \frac{1}{4} \) and \(-3\), along with \(x = 2\) which we identified earlier.These roots are the solutions to the original polynomial equation \(4x^3 + 3x^2 - 25x + 6 = 0\). Thus, establishing a systematic approach to solving polynomial equations efficiently helps in mastering these mathematical concepts.