Find all of the zeros of If f(x) = x3 + 13x2 + 32x + 20 and f(-10)=0
pls help with this problem !

Step-by-step explanation:

Well you have one of the roots -- -10. Just by looking at it, -1 would seem to be another x intercept.

f(x) = (-1)^3 + 13(-1)^2 + 32(-1) + 20

f(x) = -1 + 13 - 32 + 20

f(x) = 0

So - 1 is also a root.

You should try -2

-2 * -10 - 1 = - 20

f(-2) = (-2)^3 + 13(-2)^2 + 32(-2) + 20

f(-2) =  -8 + 13 * 4 - 32(2) + 20

f(-2) = - 8 + 52 - 64 + 20

f(-2) = -72 + 72

f(-2) = 0

The zeros are

x = -10

x = -1

x = -2

36 is the final answer! : )

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}