Is (x − 4) a factor of f(x) = x3 − 2x2 + 5x + 1? use either the remainder theorem or the factor theorem to explain your reasoning.

ANSWER

Not a factor

EXPLANATION

The given polynomial function is:

According to the remainder theorem,if (x-4) is a factor of f(x), then f(4)=0.

We substitute x=4 into the function to obtain;

Since the remainder is not zero, x-4 is not a factor.

  No, (x - 4) is NOT a factor of the polynomial f(x).

Step-by-step explanation:  We are given to check whether (x − 4) a factor of the following polynomial :

We will be using factor theorem states as follows :

FACTOR THEOREM :  If (x - a) is a factor of the polynomial p(x), then p(a) = 0.

Now, substituting x = 4 in equation (i), we get

Since f(4) ≠ 0, so (x - 4) is NOT a factor of the polynomial f(x).

Thus, (x - 4) is not a factor of the polynomial f(x).

∠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.

step-by-step explanation:

hans did not square the 40-unit side; he merely multiplied it by 2.   incorrect.