SOLUTION: Find all real and complex zeros of the polynomial, and write p(x) as the product of linear factors. P(x)=x^4-9x^2+4x+12

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Find all real and complex zeros of the polynomial, and write p(x) as the product of linear factors. P(x)=x^4-9x^2+4x+12      Log On


   

Question 1050510: Find all real and complex zeros of the polynomial, and write p(x) as the product of linear factors.
P(x)=x^4-9x^2+4x+12
Answer by ikleyn(52778) About Me  (Show Source):
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Find all real and complex zeros of the polynomial, and write p(x) as the product of linear factors.
P(x)=x^4-9x^2+4x+12
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Apply grouping to factor P(x):

x%5E4-9x%5E2%2B4x%2B12 = %28x%5E4-9x%5E2%29%2B%284x%2B12%29 = x%5E2%28x%5E2-9%29+%2B+4%2A%28x%2B3%29 = x%5E2%2A%28x%2B3%29%2A%28x-3%29+%2B+4%2A%28x%2B3%29 = %28x%2B3%29%2A%28x%5E2%2A%28x-3%29%2B4%29 = %28x%2B3%29%2A%28x%5E3-3x%5E2%2B4%29. 

So, one root is x = -3.

Next, transform and factor the polynomial Q(x) = x^3 - 3x^2 + 4.
Add and distract x^2 and then apply grouping, again:

x%5E3+-+3x%5E2+%2B+4 = %28x%5E3+%2B+x%5E2%29+-+%284x%5E2+-4%29 = x%5E2%2A%28x%2B1%29+-+4%2A%28x%5E2-1%29 = x%5E2%2A%28x%2B1%29+-+4%2A%28x%2B1%29%2A%28x-1%29 = %28x%2B1%29%2A%28x%5E2-4%29 = %28x%2B1%29%2A%28x%2B2%29%2A%28x-2%29.

The roots of the polynomial Q(x) are -1, -2, 2.

Answer. The roots of the polynomial P(x) are -3, -2, -1, and 2.