Factoring Polynomials - PSAT Math
Card 0 of 105
Factor 9_x_2 + 12_x_ + 4.
Factor 9_x_2 + 12_x_ + 4.
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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.
What is a possible value for x in x2 – 12x + 36 = 0 ?
What is a possible value for x in x2 – 12x + 36 = 0 ?
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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
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?
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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.
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
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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
Let 
and 
be integers, such that 
. If 
and 
, then what is 
?
Let
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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.
Solve for x:
$2x^2$-4=3 +5
Solve for x:
$2x^2$-4=3 +5
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$2x^2$-4=3 +5
First, add 4 to both sides:
Divide both sides by 2:
$2x^2$-4=3 +5
First, add 4 to both sides:
Divide both sides by 2:
How many negative solutions are there to the equation below?
How many negative solutions are there to the equation below?
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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.
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)
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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
Completely factor the following expression:
Completely factor the following expression:
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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
Factor the following variable
(x2 + 18x + 72)
Factor the following variable
(x2 + 18x + 72)
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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)
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
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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.
Factor and simplify:
$$\frac{64y^{2}$$ - 16}{8y + 4}
Factor and simplify:
$$\frac{64y^{2}$$ - 16}{8y + 4}
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$64y^{2}$ - 16 is a difference of squares.
The difference of squares formula is $a^{2}$ - $b^{2}$ = (a - b)(a + b).
Therefore, $$\frac{64y^{2}$$ - 16}{8y + 4} = $\frac{(8y + 4)(8y - 4)}{8y + 4}$ = 8y - 4.
$64y^{2}$ - 16 is a difference of squares.
The difference of squares formula is $a^{2}$ - $b^{2}$ = (a - b)(a + b).
Therefore, $$\frac{64y^{2}$$ - 16}{8y + 4} = $\frac{(8y + 4)(8y - 4)}{8y + 4}$ = 8y - 4.
If $$\frac{x^{2}$$-9}{x+3}=5 , and xneq -3 , what is the value of x?
If $$\frac{x^{2}$$-9}{x+3}=5 , and xneq -3 , what is the value of x?
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The numerator on the left can be factored so the expression becomes $\frac{left ( x+3 right )times left ( x-3 right )}{left ( x+3 right )}$=5, which can be simplified to left ( x-3 right )=5
Then you can solve for x by adding 3 to both sides of the equation, so x=8
The numerator on the left can be factored so the expression becomes $\frac{left ( x+3 right )times left ( x-3 right )}{left ( x+3 right )}$=5, which can be simplified to left ( x-3 right )=5
Then you can solve for x by adding 3 to both sides of the equation, so x=8
Factor:
$-12x^2$+27
Factor:
$-12x^2$+27
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We can first factor out -3:
$-3(4x^{2}$-9)
This factors further because there is a difference of squares:
-3(2x+3)(2x-3)
We can first factor out -3:
$-3(4x^{2}$-9)
This factors further because there is a difference of squares:
-3(2x+3)(2x-3)
Solve for x:
$x^2$+3x+2=0
Solve for x:
$x^2$+3x+2=0
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First, factor.
$x^2$+3x+2=(x+2)(x+1)=0
Set each factor equal to 0
x+2=0; x=-2
x+1=0; x=-1
Therefore,
x=-2 or-1
First, factor.
$x^2$+3x+2=(x+2)(x+1)=0
Set each factor equal to 0
x+2=0; x=-2
x+1=0; x=-1
Therefore,
x=-2 or-1
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
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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
What is a possible value for x in x2 – 12x + 36 = 0 ?
What is a possible value for x in x2 – 12x + 36 = 0 ?
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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
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?
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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.
Let and be integers, such that . If and , then what is ?
Let
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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.
Solve for x:
$2x^2$-4=3 +5
Solve for x:
$2x^2$-4=3 +5
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$2x^2$-4=3 +5
First, add 4 to both sides:
Divide both sides by 2:
$2x^2$-4=3 +5
First, add 4 to both sides:
Divide both sides by 2: