Solve the equation x^3-13x^2+47x-35=0 given that 1 is a zero of f(x)= x^3-13x^2+47x-35. The solution set is

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x = {1 , 5 , 7}

Step-by-step explanation:

We have x^3 - 13x^2 + 47x - 35 = 0.

Taking -13x^2, we can split it up into -x^2 - 12x^2

Taking 47x, we can split it up into 12x + 35x

Putting these into the original equation, we have:

x^3 - x^2 - 12x^2 + 12x + 35x - 35 = 0

From this you can see that we have 3 pairs of terms that we can easily factor. From the first pair we can factor x^2, from the second we can factor -12x, and from the third one we can factor 35.

1. (x^3 - x^2)  = x^2 (x - 1)

2. (- 12x^2 + 12x) = -12x (x - 1)

3. (35x - 35) = 35 (x - 1)

Putting them back into the equation we have:

x^2 (x - 1) - 12x (x - 1) + 35 (x - 1) = 0

               (x^2 - 12x + 35) (x - 1) = 0

To find the roots, we can set (x - 1) = 0 and  (x^2 - 12x + 35) = 0.

x - 1 = 0

    x = 1 (First root)

x^2 - 12x + 35 = 0

Factoring using the Sum-Product Rule:

Sum of -12, Product of 35 --> -7 and -5

x^2 - 12x + 35 = 0

  (x - 7) (x - 5) = 0

We can see that x = 7 and x = 5.

So the solution set it x = {1 , 5 , 7}

answer: about 22.67

step-by-step explanation: