Answer :
To solve the equation [tex]\(x^3 - 13x^2 + 47x - 35 = 0\)[/tex] given that 1 is a zero of the polynomial, follow these steps:
1. Understand the Problem: We know that for a polynomial [tex]\(f(x)\)[/tex], if 1 is a root, then [tex]\(x - 1\)[/tex] is a factor of the polynomial [tex]\(x^3 - 13x^2 + 47x - 35\)[/tex].
2. Polynomial Division: Since [tex]\(x - 1\)[/tex] is a factor, we can divide the original polynomial by [tex]\(x - 1\)[/tex] to find the other factor.
- Perform the division to get the quotient, [tex]\(x^2 - 12x + 35\)[/tex]. The remainder should be zero since 1 is a root.
3. Solve the Quadratic Equation:
- Now, solve the quadratic equation [tex]\(x^2 - 12x + 35 = 0\)[/tex] to find its roots. For a quadratic equation of the form [tex]\(ax^2 + bx + c = 0\)[/tex], the quadratic formula is:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
- In this case, [tex]\(a = 1\)[/tex], [tex]\(b = -12\)[/tex], and [tex]\(c = 35\)[/tex]. Plug these values into the formula to solve for [tex]\(x\)[/tex]:
[tex]\[
x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \cdot 1 \cdot 35}}{2 \cdot 1}
\][/tex]
[tex]\[
x = \frac{12 \pm \sqrt{144 - 140}}{2}
\][/tex]
[tex]\[
x = \frac{12 \pm \sqrt{4}}{2}
\][/tex]
[tex]\[
x = \frac{12 \pm 2}{2}
\][/tex]
- This gives us the roots [tex]\(x = \frac{14}{2} = 7\)[/tex] and [tex]\(x = \frac{10}{2} = 5\)[/tex].
4. List All Roots: The original polynomial [tex]\(x^3 - 13x^2 + 47x - 35\)[/tex] has the roots derived from [tex]\(x - 1\)[/tex] and the quotient:
- Thus, the roots are 1, 5, and 7.
In conclusion, the solutions to the equation [tex]\(x^3 - 13x^2 + 47x - 35 = 0\)[/tex] are [tex]\(x = 1\)[/tex], [tex]\(x = 5\)[/tex], and [tex]\(x = 7\)[/tex].
1. Understand the Problem: We know that for a polynomial [tex]\(f(x)\)[/tex], if 1 is a root, then [tex]\(x - 1\)[/tex] is a factor of the polynomial [tex]\(x^3 - 13x^2 + 47x - 35\)[/tex].
2. Polynomial Division: Since [tex]\(x - 1\)[/tex] is a factor, we can divide the original polynomial by [tex]\(x - 1\)[/tex] to find the other factor.
- Perform the division to get the quotient, [tex]\(x^2 - 12x + 35\)[/tex]. The remainder should be zero since 1 is a root.
3. Solve the Quadratic Equation:
- Now, solve the quadratic equation [tex]\(x^2 - 12x + 35 = 0\)[/tex] to find its roots. For a quadratic equation of the form [tex]\(ax^2 + bx + c = 0\)[/tex], the quadratic formula is:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
- In this case, [tex]\(a = 1\)[/tex], [tex]\(b = -12\)[/tex], and [tex]\(c = 35\)[/tex]. Plug these values into the formula to solve for [tex]\(x\)[/tex]:
[tex]\[
x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \cdot 1 \cdot 35}}{2 \cdot 1}
\][/tex]
[tex]\[
x = \frac{12 \pm \sqrt{144 - 140}}{2}
\][/tex]
[tex]\[
x = \frac{12 \pm \sqrt{4}}{2}
\][/tex]
[tex]\[
x = \frac{12 \pm 2}{2}
\][/tex]
- This gives us the roots [tex]\(x = \frac{14}{2} = 7\)[/tex] and [tex]\(x = \frac{10}{2} = 5\)[/tex].
4. List All Roots: The original polynomial [tex]\(x^3 - 13x^2 + 47x - 35\)[/tex] has the roots derived from [tex]\(x - 1\)[/tex] and the quotient:
- Thus, the roots are 1, 5, and 7.
In conclusion, the solutions to the equation [tex]\(x^3 - 13x^2 + 47x - 35 = 0\)[/tex] are [tex]\(x = 1\)[/tex], [tex]\(x = 5\)[/tex], and [tex]\(x = 7\)[/tex].
