Can someone factor this for me?
y=x^4-4x^3+16
i could only get up to:
y=(x-2)(x^3-2x^2-4x-8)
That's as far as you can go using rational numbers
(I ran x^3-2x^2-4x-8 = 0 through my "cubic equation solver and got one messy real root and 2 complex roots)
http://www.1728.com/cubic.htm
would there be another way to find the x intercepts? Cause that's what i was trying to achieve
you could sketch it and see appr. where it crosses the x-axis.
let f(x) = x^3 - 2x^2 - 4x - 8
If there are any rational factors, then test for
f(x) = 0 using x = ±1, ±2,±4, and ±8
none work, so there are no rational factors.
To factor the expression y = x^4 - 4x^3 + 16 completely, we can follow these steps:
Step 1: Look for any common factors among the terms. In this case, there are no common factors.
Step 2: Try to apply factoring techniques like factoring by grouping or using special factoring formulas. In this case, those techniques may not apply directly.
Step 3: Check for any possible rational roots using the rational root theorem. The rational root theorem states that if a rational number p/q is a root of the polynomial, then p must be a factor of the constant term (in this case, 16) and q must be a factor of the leading coefficient (which is 1). The factors of 16 are: ±1, ±2, ±4, ±8, ±16, and the factors of 1 are ±1.
Let's test each possible rational root by substituting them into the polynomial expression and see if any gives us a result of 0.
Substituting x = 1:
y = (1)^4 - 4(1)^3 + 16 = 1 - 4 + 16 = 13
Since y ≠ 0, x = 1 is not a root.
Substituting x = -1:
y = (-1)^4 - 4(-1)^3 + 16 = 1 + 4 + 16 = 21
Again, y ≠ 0, x = -1 is not a root.
Substituting x = 2:
y = (2)^4 - 4(2)^3 + 16 = 16 - 32 + 16 = 0
Finally, we find that x = 2 is a root, and we can factor out (x - 2) from the polynomial expression.
So, we have:
y = (x - 2)(x^3 - 2x^2 - 4x - 8)
Now, the cubic polynomial x^3 - 2x^2 - 4x - 8 needs further factoring. At this point, we can try applying factoring by grouping or other techniques to see if it can be factored more.