Factor completely 3x^5-27x^4+21x^3
factor out 3x^3
3x^3 (x^2 -9x+7)
Thank you so much
To factor the expression 3x^5 - 27x^4 + 21x^3 completely, we can begin by looking for the greatest common factor (GCF) of the terms. In this case, the GCF would be 3x^3, as it divides each term evenly.
Factoring out the GCF, we get:
3x^3(x^2 - 9x + 7)
Now, let's focus on factoring the quadratic expression (x^2 - 9x + 7). Since it doesn't seem to be easily factored, we can use the quadratic formula or complete the square method to find its factors. Let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = -9, and c = 7. Plugging these values into the formula, we have:
x = (-(-9) ± √((-9)^2 - 4(1)(7))) / (2(1))
x = (9 ± √(81 - 28)) / 2
x = (9 ± √53) / 2
Using the quadratic formula, we find that the factors of the quadratic expression are: (x - [(9 + √53) / 2]) and (x - [(9 - √53) / 2]).
Therefore, the completely factored form of the expression 3x^5 - 27x^4 + 21x^3 is:
3x^3(x - [(9 + √53) / 2])(x - [(9 - √53) / 2])