Factor x^2-3x+9

Problem

x2−3*x+9

Solution

1. Identify the coefficients of the quadratic expression a*x2+b*x+c where a=1 b=−3 and c=9

2. Check for two numbers that multiply to a*c=9 and add to b=−3 The pairs of factors for 9 are (1,9) (−1,−9) and (3,3) (−3,−3) None of these pairs sum to −3

3. Evaluate the discriminant D=b2−4*a*c to determine if the polynomial can be factored over the real numbers.

D=(−3)2−4*(1)*(9)

D=9−36

D=−27

4. Conclude that since the discriminant is negative, the quadratic has no real roots and cannot be factored into linear factors with real coefficients. It is considered irreducible over the real numbers.

5. Apply the quadratic formula x=(−b±√(,D))/(2*a) if factoring over the complex numbers is required.

x=(−(−3)±√(,−27))/(2*(1))

x=(3±3*i√(,3))/2

6. Write the factors using the complex roots (x_1) and (x_2) in the form (x−(x_1))*(x−(x_2))

Final Answer

x2−3*x+9=(x−(3+3*i√(,3))/2)*(x−(3−3*i√(,3))/2)

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