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.

First, add 4 to both sides:

Divide both sides by 2:

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. 

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.

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

By the Remainder Theorem, if a polynomial  is divided by a binomial , the remainder is . 

Let . Setting , if  is divided by , the remainder is , which can be evaluated by setting  in the definition of  and evaluating:

Call 

By the Rational Zeroes Theorem, since  has only integer coefficients, any rational solution of  must be a factor of 54 divided by a factor of 1 - positive or negative. 54 has as its factors 1, 2, 3, 6, 9, 18, 27 , 54; 1 has only itself as a factor. Therefore, the rational solutions of  must be chosen from this set:

.

By the Factor Theorem, a polynomial  is divisible by  if and only if  - that is, if  is a zero. By the preceding result, we can immediately eliminate  and  as factors, since 12 and 16 have been eliminated as possible zeroes.

Of the three remaining choices, we can demonstrate that  is the factor by evaluating :

, so  is a factor. 

Of the remaining two choices,  and , both can be proved to not be factors by showing that  and  are both nonzero:

, so  is not a factor.

, so  is not a factor. 

By the Remainder Theorem, if a polynomial  is divided by a binomial , the remainder is . 

Let . Setting  (since  ), if  is divided by , the remainder is , which can be evaluated by setting  in the definition of  and evaluating: