How to divide polynomials - PSAT Math
Card 0 of 42
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
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
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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
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
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?
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?
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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.
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.
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If 3 less than 15 is equal to 2x, then 24/x must be greater than
If 3 less than 15 is equal to 2x, then 24/x must be greater than
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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.
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.
Simplify: 
Simplify:
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Cancel by subtracting the exponents of like terms:
Cancel by subtracting the exponents of like terms:
What is the remainder when the polynomial 
is divided by 
?
What is the remainder when the polynomial
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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 
by 
is the dividend evaluated at 
, which is
By the remainder theorem, if a polynomial
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?
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?
Tap to see back →
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.
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.
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Simplify:
Simplify:
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Cancel by subtracting the exponents of like terms:
Cancel by subtracting the exponents of like terms:
If 3 less than 15 is equal to 2x, then 24/x must be greater than
If 3 less than 15 is equal to 2x, then 24/x must be greater than
Tap to see back →
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.
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.
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
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
Tap to see back →
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
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 is divided by ?
What is the remainder when the polynomial
Tap to see back →
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 by is the dividend evaluated at , which is
By the remainder theorem, if a polynomial
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
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
Tap to see back →
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
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
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?
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?
Tap to see back →
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.
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.
Tap to see back →
If 3 less than 15 is equal to 2x, then 24/x must be greater than
If 3 less than 15 is equal to 2x, then 24/x must be greater than
Tap to see back →
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.
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.
Simplify:
Simplify:
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Cancel by subtracting the exponents of like terms:
Cancel by subtracting the exponents of like terms:
What is the remainder when the polynomial is divided by ?
What is the remainder when the polynomial
Tap to see back →
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 by is the dividend evaluated at , which is
By the remainder theorem, if a polynomial
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
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
Tap to see back →
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
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
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?
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?
Tap to see back →
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.
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.