SOLUTION: factor completely 3x^5-12x^4+21x^3

Question 350163: factor completely 3x^5-12x^4+21x^3
Answer by CharlesG2(834) (Show Source): You can put this solution on YOUR website!
factor completely 3x^5-12x^4+21x^3

find the greatest common factor of each of the 3 terms of this trinomial (polynomial with 3 terms) and factor it out

that is 3x^3 since the coefficients 3, -12, and 21 are all divisible by 3, and x^5, x^4, and x^3 are all divisible by x^3

3x^3(x^2 - 4x + 7)
now see if x^2 - 4x + 7 can be factored

set the trinomial equal to 0, now it is a quadratic equation of
the form ax^2 + bx + c = 0, where a = 1, b = -4, and c = 7
use the discriminant of the quadratic equation to determine if there is solutions among real numbers (any number that is not imaginary, does not involve the imaginary number i, where i is square root of -1)

disciminant b^2 - 4ac = -4^2 - 4 * 1 * 7 = 16 - 28 = -12
discriminant is negative, no solutions among real numbers
x^2 - 4x + 7 can not be factored using real numbers

3x^5-12x^4+21x^3 factored completely is 3x^3(x^2 - 4x + 7) without going into complex numbers (numbers of form a + bi where a and b are real numbers and i is the square root of -1)

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