How to factor a polynomial - PSAT Math
Card 0 of 56
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
Compare your answer with the correct one above
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.
Compare your answer with the correct one above
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
Compare your answer with the correct one above
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.
Compare your answer with the correct one above
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:
Compare your answer with the correct one above
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.
Compare your answer with the correct one above
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
Compare your answer with the correct one above
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
Compare your answer with the correct one above
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
Compare your answer with the correct one above
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.
Compare your answer with the correct one above
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
Compare your answer with the correct one above
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.
Compare your answer with the correct one above
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:
Compare your answer with the correct one above
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.
Compare your answer with the correct one above
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
Compare your answer with the correct one above
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
Compare your answer with the correct one above
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
Compare your answer with the correct one above
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.
Compare your answer with the correct one above
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
Compare your answer with the correct one above
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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