PSAT Math - How to divide polynomials

If a positive integer a is divided by 7, the remainder is 4. What is the remainder if 3_a_ + 5 is divided by 3?

Explanation

The best way to solve this problem is to plug in an appropriate value for a. For example, plug-in 11 for a because 11 divided by 7 will give us a remainder of 4.

Then 3_a + 5_, where a = 11, gives us 38. Then 38 divided by 3 gives a remainder of 2.

The algebra method is as follows:

a divided by 7 gives us some positive integer b, with a remainder of 4.

Thus,

a / 7 = b 4/7

a / 7 = (7_b +_ 4) / 7

a = (7_b_ + 4)

then 3_a + 5 =_ 3 (7_b_ + 4) + 5

(3_a_+5)/3 = \[3(7_b_ + 4) + 5\] / 3

= (7_b_ + 4) + 5/3

The first half of this expression (7_b_ + 4) is a positive integer, but the second half of this expression (5/3) gives us a remainder of 2.

If 3 less than 15 is equal to 2x, then 24/x must be greater than

Explanation

Set up an equation for the sentence: 15 – 3 = 2x and solve for x. X equals 6. If you plug in 6 for x in the expression 24/x, you get24/6 = 4. 4 is only choice greater than a.

Polydivision1

100

Explanation

Polydivision2

Polydivision4

Given a♦b = (a+b)/(a-b) and b♦a = (b+a)/(b-a), which of the following statement(s) is(are) true:

I. a♦b = -(b♦a)

II. (a♦b)(b♦a) = (a♦b)2

III. a♦b + b♦a = 0

I only

I and II

II & III

I and III

I, II and III

Explanation

Notice that - (a-b) = b-a, so statement I & III are true after substituting the expression. Substitute the expression for statement II gives ((a+b)/(a-b))((a+b)/(b-a))=((a+b)(b+a))/((-1)(a-b)(a-b))=-1 〖(a+b)〗2/〖(a-b)〗2 =-((a+b)/(a-b))2 = -(a♦b)2 ≠ (a♦b)2

What is the remainder when the polynomial $x^{4} - 5x^{3} + 3x - 17$ is divided by ?

Explanation

By the remainder theorem, if a polynomial is divided by the linear binomial , the remainder is - that is, the polynomial evaluated at . The remainder of dividing $x^{4} - 5x^{3} + 3x - 17$ by is the dividend evaluated at , which is