Will mark !
given the following equation, x^4 - x^3 - 3x - 2. use synthetic division to test the following 3 points and show which one is a root. must show synthetic division for all 3 of the following test points. test these 3 possible root points also called possible zeros using synthetic division: (x, y) = (-1, 0) (-3, 0) (2, 0).
show your !

Answer x^2-5=0

x^2= 5

x=+ v/5

The answer to your question is:

(-1,0)           it's not a root

(-3,0)          it's not a root

(2,0)           it's a root

Step-by-step explanation:

                                            1        -1        0        -3        -2             -1

                                                      -1       2         -2        5

                                             1       -2       2        -5         3    it's not a root

                                            1        -1        0        -3        -2             -3

                                                      -3      -12      36       99  

                                            1       -4       -12       33       97     It is not a root    

                                              1        -1        0        -3        -2             2

                                                        2        2         4          2

                                              1          1        2          1          0    It's a root

pooopoopopoopop

step-by-step explanati∵∵∴∴on:

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.