Question

if (f(x)=x^{3}-2x^{2}+7x - 5) and (x - 1) is a factor of (f(x)), then find all of the zeros of (f(x)) algebraically.

Explanation:

Step1: Use factor theorem

If \(x - a\) is a factor of \(f(x)\), then \(f(a)=0\). Here \(a = 1\).

Step2: Perform synthetic division

Divide \(f(x)\) by \(x - 1\) to get the quotient polynomial.

Step3: Solve quotient polynomial

Use quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for the quadratic quotient \(ax^{2}+bx + c\) to find its roots.

Step4: Combine roots

Combine the root from \(x - 1\) and the roots of the quotient polynomial to get all roots of \(f(x)\).

Answer:

Since \(x - 1\) is a factor of \(f(x)=x^{3}-2x^{2}+7x - 5\), we can use polynomial long - division or synthetic division.
Using synthetic division:
The coefficients of \(f(x)\) are \(1,-2,7,-5\).

1. Perform synthetic division with \(x = 1\):

• Bring down the first coefficient \(1\):
• Multiply \(1\times1 = 1\), add to the second coefficient: \(-2 + 1=-1\).
• Multiply \(1\times(-1)=-1\), add to the third coefficient: \(7+( - 1)=6\).
• Multiply \(1\times6 = 6\), add to the fourth coefficient: \(-5 + 6 = 1\).
• The quotient is \(x^{2}-x + 6\) and the remainder is \(1\). But there is a mistake above, let's use polynomial long - division correctly.
• Divide \(x^{3}-2x^{2}+7x - 5\) by \(x - 1\):
• \(x^{3}\div x=x^{2}\), then \((x - 1)\times x^{2}=x^{3}-x^{2}\), and \((x^{3}-2x^{2}+7x - 5)-(x^{3}-x^{2})=-x^{2}+7x - 5\).
• \(-x^{2}\div x=-x\), then \((x - 1)\times(-x)=-x^{2}+x\), and \((-x^{2}+7x - 5)-(-x^{2}+x)=6x - 5\).
• \(6x\div x = 6\), then \((x - 1)\times6=6x-6\), and \((6x - 5)-(6x - 6)=1\).
• The quotient is \(x^{2}-x + 6\) and remainder \(1\). There is an error in the problem setup as if \(x - 1\) is a factor the remainder should be \(0\). Let's assume the polynomial is \(f(x)=x^{3}-2x^{2}+7x - 6\).
• Using synthetic division with \(x = 1\):
• Bring down \(1\), \(1\times1 = 1\), \(-2+1=-1\), \(1\times(-1)=-1\), \(7+( - 1)=6\), \(1\times6 = 6\), \(-6 + 6=0\).
• The quotient is \(x^{2}-x + 6\).
• Now, find the roots of \(x^{2}-x + 6\) using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c\). Here \(a = 1\), \(b=-1\), \(c = 6\).
• \(\Delta=b^{2}-4ac=(-1)^{2}-4\times1\times6=1 - 24=-23\lt0\).
• The roots of \(x^{2}-x + 6\) are \(x=\frac{1\pm\sqrt{-23}}{2}=\frac{1\pm i\sqrt{23}}{2}\).
• The root of \(x - 1\) is \(x = 1\).
• The roots of \(f(x)=x^{3}-2x^{2}+7x - 6\) are \(x = 1,\frac{1 + i\sqrt{23}}{2},\frac{1 - i\sqrt{23}}{2}\).