How to factor a polynomial - PSAT Math

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

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 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.

$2x^2$-4=3 +5

First, add 4 to both sides:

Divide both sides by 2:

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.

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.

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

.

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

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.

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

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.

$2x^2$-4=3 +5

First, add 4 to both sides:

Divide both sides by 2:

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.

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.

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

.

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

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.

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

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.