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Answer :
To solve the equation [tex]\(45x^3 + 69x^2 - 13x - 5 = 0\)[/tex] given that [tex]\(-\frac{5}{3}\)[/tex] is a zero, we can follow these steps:
1. Verify the Given Root:
Since [tex]\(-\frac{5}{3}\)[/tex] is a zero of the polynomial [tex]\(f(x) = 45x^3 + 69x^2 - 13x - 5\)[/tex], it means that the polynomial can be divided by the factor [tex]\((x + \frac{5}{3})\)[/tex] without a remainder.
2. Factor the Polynomial:
We know that one factor is [tex]\((3x + 5)\)[/tex] because [tex]\(-\frac{5}{3}\)[/tex] is a root. We can factor the cubic polynomial by dividing it by this factor to find the remaining quadratic polynomial.
3. Perform Polynomial Division:
Upon dividing [tex]\(45x^3 + 69x^2 - 13x - 5\)[/tex] by [tex]\((3x + 5)\)[/tex], we get another factor, which is a quadratic polynomial. Let's represent this resulting polynomial as:
[tex]\[
45x^3 + 69x^2 - 13x - 5 = (3x + 5)(Ax^2 + Bx + C)
\][/tex]
Upon division, the quotient is [tex]\(45x^2 - 6x - 3\)[/tex].
4. Solve the Quadratic Factor:
Now, solve the quadratic equation [tex]\(45x^2 - 6x - 3 = 0\)[/tex] to find the other roots. Using the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
For the quadratic [tex]\(45x^2 - 6x - 3\)[/tex], we have [tex]\(a = 45\)[/tex], [tex]\(b = -6\)[/tex], and [tex]\(c = -3\)[/tex]. Calculate:
[tex]\[
x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 45 \cdot (-3)}}{2 \cdot 45}
\][/tex]
[tex]\[
x = \frac{6 \pm \sqrt{36 + 540}}{90}
\][/tex]
[tex]\[
x = \frac{6 \pm \sqrt{576}}{90}
\][/tex]
[tex]\[
x = \frac{6 \pm 24}{90}
\][/tex]
Solutions are:
- [tex]\(x = \frac{6 + 24}{90} = \frac{30}{90} = \frac{1}{3}\)[/tex]
- [tex]\(x = \frac{6 - 24}{90} = \frac{-18}{90} = -\frac{1}{5}\)[/tex]
5. Conclusion:
The solutions of the equation [tex]\(45x^3 + 69x^2 - 13x - 5 = 0\)[/tex] are:
[tex]\[
x = -\frac{5}{3}, \, x = \frac{1}{3}, \, \text{and} \, x = -\frac{1}{5}
\][/tex]
These roots satisfy the original cubic polynomial equation, completing the solution.
1. Verify the Given Root:
Since [tex]\(-\frac{5}{3}\)[/tex] is a zero of the polynomial [tex]\(f(x) = 45x^3 + 69x^2 - 13x - 5\)[/tex], it means that the polynomial can be divided by the factor [tex]\((x + \frac{5}{3})\)[/tex] without a remainder.
2. Factor the Polynomial:
We know that one factor is [tex]\((3x + 5)\)[/tex] because [tex]\(-\frac{5}{3}\)[/tex] is a root. We can factor the cubic polynomial by dividing it by this factor to find the remaining quadratic polynomial.
3. Perform Polynomial Division:
Upon dividing [tex]\(45x^3 + 69x^2 - 13x - 5\)[/tex] by [tex]\((3x + 5)\)[/tex], we get another factor, which is a quadratic polynomial. Let's represent this resulting polynomial as:
[tex]\[
45x^3 + 69x^2 - 13x - 5 = (3x + 5)(Ax^2 + Bx + C)
\][/tex]
Upon division, the quotient is [tex]\(45x^2 - 6x - 3\)[/tex].
4. Solve the Quadratic Factor:
Now, solve the quadratic equation [tex]\(45x^2 - 6x - 3 = 0\)[/tex] to find the other roots. Using the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
For the quadratic [tex]\(45x^2 - 6x - 3\)[/tex], we have [tex]\(a = 45\)[/tex], [tex]\(b = -6\)[/tex], and [tex]\(c = -3\)[/tex]. Calculate:
[tex]\[
x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 45 \cdot (-3)}}{2 \cdot 45}
\][/tex]
[tex]\[
x = \frac{6 \pm \sqrt{36 + 540}}{90}
\][/tex]
[tex]\[
x = \frac{6 \pm \sqrt{576}}{90}
\][/tex]
[tex]\[
x = \frac{6 \pm 24}{90}
\][/tex]
Solutions are:
- [tex]\(x = \frac{6 + 24}{90} = \frac{30}{90} = \frac{1}{3}\)[/tex]
- [tex]\(x = \frac{6 - 24}{90} = \frac{-18}{90} = -\frac{1}{5}\)[/tex]
5. Conclusion:
The solutions of the equation [tex]\(45x^3 + 69x^2 - 13x - 5 = 0\)[/tex] are:
[tex]\[
x = -\frac{5}{3}, \, x = \frac{1}{3}, \, \text{and} \, x = -\frac{1}{5}
\][/tex]
These roots satisfy the original cubic polynomial equation, completing the solution.
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