Algebra - PSAT Math
Card 0 of 5075
What is a possible value for x in x2 – 12x + 36 = 0 ?
What is a possible value for x in x2 – 12x + 36 = 0 ?
You need to factor to find the possible values for x. You need to fill in the blanks with two numbers with a sum of -12 and a product of 36. In both sets of parenthesis, you know you will be subtracting since a negative times a negative is a positive and a negative plus a negative is a negative
(x –__)(x –__).
You should realize that 6 fits into both blanks.
You must now set each set of parenthesis equal to 0.
x – 6 = 0; x – 6 = 0
Solve both equations: x = 6
You need to factor to find the possible values for x. You need to fill in the blanks with two numbers with a sum of -12 and a product of 36. In both sets of parenthesis, you know you will be subtracting since a negative times a negative is a positive and a negative plus a negative is a negative
(x –__)(x –__).
You should realize that 6 fits into both blanks.
You must now set each set of parenthesis equal to 0.
x – 6 = 0; x – 6 = 0
Solve both equations: x = 6
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If r and t are constants and x2 +rx +6=(x+2)(x+t), what is the value of r?
If r and t are constants and x2 +rx +6=(x+2)(x+t), what is the value of r?
We first expand the right hand side as x2+2x+tx+2t and factor out the x terms to get x2+(2+t)x+2t. Next we set this equal to the original left hand side to get x2+rx +6=x2+(2+t)x+2t, and then we subtract x2 from each side to get rx +6=(2+t)x+2t. Since the coefficients of the x terms on each side must be equal, and the constant terms on each side must be equal, we find that r=2+t and 6=2t, so t is equal to 3 and r is equal to 5.
We first expand the right hand side as x2+2x+tx+2t and factor out the x terms to get x2+(2+t)x+2t. Next we set this equal to the original left hand side to get x2+rx +6=x2+(2+t)x+2t, and then we subtract x2 from each side to get rx +6=(2+t)x+2t. Since the coefficients of the x terms on each side must be equal, and the constant terms on each side must be equal, we find that r=2+t and 6=2t, so t is equal to 3 and r is equal to 5.
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2x + 3y = 5a + 2b (1)
3x + 2y = 4a – b (2)
Express x2 – y2 in terms of a and b
2x + 3y = 5a + 2b (1)
3x + 2y = 4a – b (2)
Express x2 – y2 in terms of a and b
Add the two equations together to yield 5x + 5y = 9a + b, then factor out 5 to get 5(x + y) = 9a + b; divide both sides by 5 to get x + y = (9a + b)/5; subtract the two equations to get x - y = -a - 3b. So, x2 – y2 = (x + y)(x – y) = (9a + b)/5 (–a – 3b) = (–\[(9a)\]2 – 28ab – \[(3b)\]2)/5
Add the two equations together to yield 5x + 5y = 9a + b, then factor out 5 to get 5(x + y) = 9a + b; divide both sides by 5 to get x + y = (9a + b)/5; subtract the two equations to get x - y = -a - 3b. So, x2 – y2 = (x + y)(x – y) = (9a + b)/5 (–a – 3b) = (–\[(9a)\]2 – 28ab – \[(3b)\]2)/5
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Let 
and 
be integers, such that 
. If 
and 
, then what is 
?
Let
We are told that x3 - y3 = 56. We can factor the left side of the equation using the formula for difference of cubes.
x3 - y3 = (x - y)(x2 + xy + y2) = 56
Since x - y = 2, we can substitute this value in for the factor x - y.
2(x2 + xy + y2) = 56
Divide both sides by 2.
x2 + xy + y2 = 28
Because we are trying to find x2 + y2, if we can get rid of xy, then we would have our answer.
We are told that 3xy = 24. If we divide both sides by 3, we see that xy = 8.
We can then substitute this value into the equation x2 + xy + y2 = 28.
x2 + 8 + y2 = 28
Subtract both sides by eight.
x2 + y2 = 20.
The answer is 20.
ALTERNATE SOLUTION:
We are told that x - y = 2 and 3xy = 24. This is a system of equations.
If we solve the first equation in terms of x, we can then substitute this into the second equation.
x - y = 2
Add y to both sides.
x = y + 2
Now we will substitute this value for x into the second equation.
3(y+2)(y) = 24
Now we can divide both sides by three.
(y+2)(y) = 8
Then we distribute.
y2 + 2y = 8
Subtract 8 from both sides.
y2 + 2y - 8 = 0
We need to factor this by thinking of two numbers that multiply to give -8 but add to give 2. These numbers are 4 and -2.
(y + 4)(y - 2) = 0
This means either y - 4 = 0, or y + 2 = 0
y = -4, or y = 2
Because x = y + 2, if y = -4, then x must be -2. Similarly, if y = 2, then x must be 4.
Let's see which combination of x and y will satisfy the final equation that we haven't used, x3 - y3 = 56.
If x = -2 and y = -4, then
(-2)3 - (-4)3 = -8 - (-64) = 56. So that means that x= -2 and y = -4 is a valid solution.
If x = 4 and y = 2, then
(4)3 - 23 = 64 - 8 = 56. So that means x = 4 and y = 2 is also a valid solution.
The final value we are asked to find is x2 + y2.
If x= -2 and y = -4, then x2 + y2 = (-2)2 + (-4)2 = 4 + 16 = 20.
If x = 4 and y = 2, then x2 + y2 = (4)2 + 22 = 16 + 4 = 20.
Thus, no matter which solution we use for x and y, x2 + y2 = 20.
The answer is 20.
We are told that x3 - y3 = 56. We can factor the left side of the equation using the formula for difference of cubes.
x3 - y3 = (x - y)(x2 + xy + y2) = 56
Since x - y = 2, we can substitute this value in for the factor x - y.
2(x2 + xy + y2) = 56
Divide both sides by 2.
x2 + xy + y2 = 28
Because we are trying to find x2 + y2, if we can get rid of xy, then we would have our answer.
We are told that 3xy = 24. If we divide both sides by 3, we see that xy = 8.
We can then substitute this value into the equation x2 + xy + y2 = 28.
x2 + 8 + y2 = 28
Subtract both sides by eight.
x2 + y2 = 20.
The answer is 20.
ALTERNATE SOLUTION:
We are told that x - y = 2 and 3xy = 24. This is a system of equations.
If we solve the first equation in terms of x, we can then substitute this into the second equation.
x - y = 2
Add y to both sides.
x = y + 2
Now we will substitute this value for x into the second equation.
3(y+2)(y) = 24
Now we can divide both sides by three.
(y+2)(y) = 8
Then we distribute.
y2 + 2y = 8
Subtract 8 from both sides.
y2 + 2y - 8 = 0
We need to factor this by thinking of two numbers that multiply to give -8 but add to give 2. These numbers are 4 and -2.
(y + 4)(y - 2) = 0
This means either y - 4 = 0, or y + 2 = 0
y = -4, or y = 2
Because x = y + 2, if y = -4, then x must be -2. Similarly, if y = 2, then x must be 4.
Let's see which combination of x and y will satisfy the final equation that we haven't used, x3 - y3 = 56.
If x = -2 and y = -4, then
(-2)3 - (-4)3 = -8 - (-64) = 56. So that means that x= -2 and y = -4 is a valid solution.
If x = 4 and y = 2, then
(4)3 - 23 = 64 - 8 = 56. So that means x = 4 and y = 2 is also a valid solution.
The final value we are asked to find is x2 + y2.
If x= -2 and y = -4, then x2 + y2 = (-2)2 + (-4)2 = 4 + 16 = 20.
If x = 4 and y = 2, then x2 + y2 = (4)2 + 22 = 16 + 4 = 20.
Thus, no matter which solution we use for x and y, x2 + y2 = 20.
The answer is 20.
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Solve for 
:
Solve for
First, add 4 to both sides:
Divide both sides by 2:
First, add 4 to both sides:
Divide both sides by 2:
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How many negative solutions are there to the equation below?
How many negative solutions are there to the equation below?
First, subtract 3 from both sides in order to obtain an equation that equals 0:
The left side can be factored. We need factors of 
that add up to 
. 
and 
work:
Set both factors equal to 0 and solve:
To solve the left equation, add 1 to both sides. To solve the right equation, subtract 3 from both sides. This yields two solutions:
Only one of these solutions is negative, so the answer is 1.
First, subtract 3 from both sides in order to obtain an equation that equals 0:
The left side can be factored. We need factors of
Set both factors equal to 0 and solve:
To solve the left equation, add 1 to both sides. To solve the right equation, subtract 3 from both sides. This yields two solutions:
Only one of these solutions is negative, so the answer is 1.
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How many of the following are prime factors of the polynomial 
?
(A) 
(B) 
(C) 
(D) 
How many of the following are prime factors of the polynomial
(A)
(B)
(C)
(D)
can be seen to fit the pattern
:
where 
can be factored as 
, so
, as the sum of squares, is a prime polynomial, so the complete factorization is
,
making 
the only prime factor, and "one" the correct choice.
where
making
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Completely factor the following expression:
Completely factor the following expression:
To begin, factor out any like terms from the expression. In this case, the term 
can be pulled out:
Next, recall the difference of squares:
Here, 
and 
.
Thus, our answer is
.
To begin, factor out any like terms from the expression. In this case, the term
Next, recall the difference of squares:
Here,
Thus, our answer is
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Factor the following variable
(x2 + 18x + 72)
Factor the following variable
(x2 + 18x + 72)
You need to find two numbers that multiply to give 72 and add up to give 18
easiest way: write the multiples of 72:
1, 72
2, 36
3, 24
4, 18
6, 12: these add up to 18
(x + 6)(x + 12)
You need to find two numbers that multiply to give 72 and add up to give 18
easiest way: write the multiples of 72:
1, 72
2, 36
3, 24
4, 18
6, 12: these add up to 18
(x + 6)(x + 12)
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When 
is factored, it can be written in the form 
, where 
, 
, 
, 
, 
, and 
are all integer constants, and 
.
What is the value of 
?
When
What is the value of
Let's try to factor x2 – y2 – z2 + 2yz.
Notice that the last three terms are very close to y2 + z2 – 2yz, which, if we rearranged them, would become y2 – 2yz+ z2. We could factor y2 – 2yz+ z2 as (y – z)2, using the general rule that p2 – 2pq + q2 = (p – q)2 .
So we want to rearrange the last three terms. Let's group them together first.
x2 + (–y2 – z2 + 2yz)
If we were to factor out a –1 from the last three terms, we would have the following:
x2 – (y2 + z2 – 2yz)
Now we can replace y2 + z2 – 2yz with (y – z)2.
x2 – (y – z)2
This expression is actually a differences of squares. In general, we can factor p2 – q2 as (p – q)(p + q). In this case, we can substitute x for p and (y – z) for q.
x2 – (y – z)2 = (x – (y – z))(x + (y – z))
Now, let's distribute the negative one in the trinomial x – (y – z)
(x – (y – z))(x + (y – z))
(x – y + z)(x + y – z)
The problem said that factoring x2 – y2 – z2 + 2yz would result in two polynomials in the form (ax + by + cz)(dx + ey + fz), where a, b, c, d, e, and f were all integers, and a > 0.
(x – y + z)(x + y – z) fits this form. This means that a = 1, b = –1, c = 1, d = 1, e = 1, and f = –1. The sum of all of these is 2.
The answer is 2.
Let's try to factor x2 – y2 – z2 + 2yz.
Notice that the last three terms are very close to y2 + z2 – 2yz, which, if we rearranged them, would become y2 – 2yz+ z2. We could factor y2 – 2yz+ z2 as (y – z)2, using the general rule that p2 – 2pq + q2 = (p – q)2 .
So we want to rearrange the last three terms. Let's group them together first.
x2 + (–y2 – z2 + 2yz)
If we were to factor out a –1 from the last three terms, we would have the following:
x2 – (y2 + z2 – 2yz)
Now we can replace y2 + z2 – 2yz with (y – z)2.
x2 – (y – z)2
This expression is actually a differences of squares. In general, we can factor p2 – q2 as (p – q)(p + q). In this case, we can substitute x for p and (y – z) for q.
x2 – (y – z)2 = (x – (y – z))(x + (y – z))
Now, let's distribute the negative one in the trinomial x – (y – z)
(x – (y – z))(x + (y – z))
(x – y + z)(x + y – z)
The problem said that factoring x2 – y2 – z2 + 2yz would result in two polynomials in the form (ax + by + cz)(dx + ey + fz), where a, b, c, d, e, and f were all integers, and a > 0.
(x – y + z)(x + y – z) fits this form. This means that a = 1, b = –1, c = 1, d = 1, e = 1, and f = –1. The sum of all of these is 2.
The answer is 2.
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Factor 9_x_2 + 12_x_ + 4.
Factor 9_x_2 + 12_x_ + 4.
Nothing common cancels at the beginning. To factor this, we need to find two numbers that multiply to 9 * 4 = 36 and sum to 12. 6 and 6 work.
So 9_x_2 + 12_x_ + 4 = 9_x_2 + 6_x_ + 6_x_ + 4
Let's look at the first two terms and last two terms separately to begin with. 9_x_2 + 6_x_ can be simplified to 3_x_(3_x_ + 2) and 6_x_ + 4 can be simplified into 2(3_x_ + 2). Putting these together gets us
9_x_2 + 12_x_ + 4
= 9_x_2 + 6_x_ + 6_x_ + 4
= 3_x_(3_x_ + 2) + 2(3_x_ + 2)
= (3_x_ + 2)(3_x_ + 2)
This is as far as we can factor.
Nothing common cancels at the beginning. To factor this, we need to find two numbers that multiply to 9 * 4 = 36 and sum to 12. 6 and 6 work.
So 9_x_2 + 12_x_ + 4 = 9_x_2 + 6_x_ + 6_x_ + 4
Let's look at the first two terms and last two terms separately to begin with. 9_x_2 + 6_x_ can be simplified to 3_x_(3_x_ + 2) and 6_x_ + 4 can be simplified into 2(3_x_ + 2). Putting these together gets us
9_x_2 + 12_x_ + 4
= 9_x_2 + 6_x_ + 6_x_ + 4
= 3_x_(3_x_ + 2) + 2(3_x_ + 2)
= (3_x_ + 2)(3_x_ + 2)
This is as far as we can factor.
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Factor and simplify:
Factor and simplify:
is a difference of squares.
The difference of squares formula is 
.
Therefore, 
= 
.
The difference of squares formula is
Therefore,
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If 
, and 
, what is the value of 
?
If
The numerator on the left can be factored so the expression becomes 
, which can be simplified to 
Then you can solve for 
by adding 3 to both sides of the equation, so 
The numerator on the left can be factored so the expression becomes
Then you can solve for
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Factor:
Factor:
We can first factor out 
:
This factors further because there is a difference of squares:
We can first factor out
This factors further because there is a difference of squares:
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Solve for x:
Solve for x:
First, factor.
Set each factor equal to 0
Therefore,
First, factor.
Set each factor equal to 0
Therefore,
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Add the polynomials.
Add the polynomials.
We can add together each of the terms of the polynomial which have the same degree for our variable. 
We can add together each of the terms of the polynomial which have the same degree for our variable.
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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
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.
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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
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
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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?
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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