All SAT Math Resources
Example Questions
Example Question #1 : How To Find The Degree Of A Polynomial
Solve each problem and decide which is the best of the choices given.
What is the degree of the following polynomial?
The degree is defined as the largest exponent in the polynomial. In this case, it is .
Example Question #2 : Polynomials
What is the degree of this polynomial?
Degree 12
Degree 6
Degree 8
Degree 10
Degree 7
Degree 8
When an exponent with a power is raised to another power, the value of the power are multiplied.
When multiplying exponents you add the powers together
The degree of a polynomial is the determined by the highest power. In this problem the highest power is 8.
Example Question #2 : Polynomials
Find the degree of the following polynomial:
The degree of a polynomial is the largest exponent on one of its variables (for a single variable), or the largest sum of exponents on variables in a single term (for multiple variables).
Here, the term with the largest exponent is , so the degree of the whole polynomial is 6.
Example Question #1 : Polynomials
If 3 less than 15 is equal to 2x, then 24/x must be greater than
5
6
4
3
3
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 get 24/6 = 4. 4 is only choice greater than a.
Example Question #2 : Polynomials
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 III
I, II and III
I and II
II & III
I and III
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
Example Question #1 : How To Divide Polynomials
If a positive integer a is divided by 7, the remainder is 4. What is the remainder if 3a + 5 is divided by 3?
5
2
6
3
4
2
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 3a + 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 = (7b + 4) / 7
a = (7b + 4)
then 3a + 5 = 3 (7b + 4) + 5
(3a+5)/3 = [3(7b + 4) + 5] / 3
= (7b + 4) + 5/3
The first half of this expression (7b + 4) is a positive integer, but the second half of this expression (5/3) gives us a remainder of 2.
Example Question #1 : Polynomials
38
42
36
45
100
42
Example Question #31 : Algebra
Simplify:
Cancel by subtracting the exponents of like terms:
Example Question #12 : Polynomials
Divide by .
It is not necessary to work a long division if you recognize as the sum of two perfect cube expressions:
A sum of cubes can be factored according to the pattern
,
so, setting ,
Therefore,
Example Question #374 : Algebra
By what expression can be multiplied to yield the product ?
Divide by by setting up a long division.
Divide the lead term of the dividend, , by that of the divisor, ; the result is
Enter that as the first term of the quotient. Multiply this by the divisor:
Subtract this from the dividend. This is shown in the figure below.
Repeat the process with the new difference:
Repeating:
The quotient - and the correct response - is .
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