please shoe me how to factor this problem b^3-4b^2-9b+36
use "grouping"
b^3-4b^2-9b+36
= b^2(b-4) - 9(b-4)
= (b-4)(b^2-9)
= (b-4)(b+3)(b-3)
x3 + 2x2 + x + 2
To factor the polynomial b^3 - 4b^2 - 9b + 36, we can try to identify any factors that can be pulled out using the method of synthetic division or by setting the polynomial equal to zero and trying different integer values for b. Let's use the second method.
1. Start by setting the polynomial equal to zero: b^3 - 4b^2 - 9b + 36 = 0.
2. Now, let's try different integer values for b to see if they make the equation true. We can start with b = 1:
(1)^3 - 4(1)^2 - 9(1) + 36 = 1 - 4 - 9 + 36 = 24
Since 24 is not equal to zero, b = 1 is not a factor.
3. Let's try b = -1:
(-1)^3 - 4(-1)^2 - 9(-1) + 36 = -1 - 4 + 9 + 36 = 40
Again, 40 is not equal to zero, so b = -1 is not a factor.
4. Next, let's try b = 2:
(2)^3 - 4(2)^2 - 9(2) + 36 = 8 - 16 - 18 + 36 = 10
Once again, 10 is not equal to zero, so b = 2 is also not a factor.
5. Finally, let's try b = -2:
(-2)^3 - 4(-2)^2 - 9(-2) + 36 = -8 - 16 + 18 + 36 = 30
Just like the previous cases, 30 is not equal to zero, so b = -2 is not a factor.
After trying multiple integer values and not finding any that make the equation true, it appears that there are no linear factors for this polynomial. Therefore, it cannot be factored further using integer values.
However, we can still factor it by using synthetic division or a graphing calculator to find any possible rational roots.