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.

\(\displaystyle 2x^2-4=3 +5\)

First, add 4 to both sides:

\(\displaystyle 2x^2=12\)

Divide both sides by 2:

\(\displaystyle x^2=6\)

\(\displaystyle x=\pm \sqrt{6}\)

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:

\(\displaystyle x^2 +2x-3=0\)

The left side can be factored. We need factors of \(\displaystyle -3\) that add up to \(\displaystyle 2\). \(\displaystyle -1\) and \(\displaystyle 3\) work:

\(\displaystyle (x-1)(x+3) = 0\)

Set both factors equal to 0 and solve:

\(\displaystyle x - 1 = 0 \qquad \mbox{ or } \qquad x+3 =0\)

To solve the left equation, add 1 to both sides. To solve the right equation, subtract 3 from both sides. This yields two solutions:

\(\displaystyle x=1 \qquad x = -3\)

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 \(\displaystyle P(x)\) is divided by a binomial \(\displaystyle x-c\), the remainder is \(\displaystyle P(c)\). 

Let \(\displaystyle P(x)= x^{5} + x^{4} - x^{3} - x^{2}\). Setting \(\displaystyle c = 2\), if \(\displaystyle P(x)\) is divided by \(\displaystyle x - 2\), the remainder is \(\displaystyle P(2)\), which can be evaluated by setting \(\displaystyle x = 2\) in the definition of \(\displaystyle P(x)\) and evaluating:

\(\displaystyle P(x)= x^{5} + x^{4} - x^{3} - x^{2}\)

\(\displaystyle P(2)= 2^{5} + 2^{4} - 2^{3} - 2^{2}\)

\(\displaystyle = 32 + 16 - 8 - 4\)

\(\displaystyle = 36\)

Call \(\displaystyle P (x) = x^{4} -6x ^{3} + 2x ^{2} - 3x - 54\)

By the Rational Zeroes Theorem, since \(\displaystyle P (x)\) has only integer coefficients, any rational solution of \(\displaystyle P (x) = 0\) 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 \(\displaystyle P (x) = 0\) must be chosen from this set:

\(\displaystyle \left \{ 1, 2, 3, 6, 9, 18, 27 , 54, -1,- 2,- 3,- 6, -9, -18, -27 ,- 54 \right \}\).

By the Factor Theorem, a polynomial \(\displaystyle P (x)\) is divisible by \(\displaystyle x- c\) if and only if \(\displaystyle P(c) = 0\) - that is, if \(\displaystyle c\) is a zero. By the preceding result, we can immediately eliminate \(\displaystyle x - 12\) and \(\displaystyle x- 16\) as factors, since 12 and 16 have been eliminated as possible zeroes.

Of the three remaining choices, we can demonstrate that \(\displaystyle x - 6\) is the factor by evaluating \(\displaystyle P(6)\):

\(\displaystyle P (x) = x^{4} -6x ^{3} + 2x ^{2} - 3x - 54\)

\(\displaystyle P (6) = 6^{4} -6 \cdot 6 ^{3} + 2 \cdot 6 ^{2} - 3 \cdot 6 - 54\)

\(\displaystyle = 1,296 -6 \cdot 216 + 2 \cdot 36 - 3 \cdot 6 - 54\)

\(\displaystyle = 1,296 -1,296 + 72 - 18 - 54\)

\(\displaystyle = 0\)

\(\displaystyle P (6) = 0\), so \(\displaystyle x - 6\) is a factor. 

Of the remaining two choices, \(\displaystyle x-3\) and \(\displaystyle x - 9\), both can be proved to not be factors by showing that \(\displaystyle P(3)\) and \(\displaystyle P (9)\) are both nonzero:

\(\displaystyle P (x) = x^{4} -6x ^{3} + 2x ^{2} - 3x - 54\)

\(\displaystyle P (3) = 3^{4} -6 \cdot 3 ^{3} + 2 \cdot 3 ^{2} - 3 \cdot 3 - 54\)

\(\displaystyle = 81 -6 \cdot 27+ 2 \cdot 9 - 3 \cdot 3 - 54\)

\(\displaystyle = 81 -162 + 18 - 9 - 54\)

\(\displaystyle =-126\)

\(\displaystyle P (3) \ne 0\), so \(\displaystyle x-3\) is not a factor.

\(\displaystyle P (x) = x^{4} -6x ^{3} + 2x ^{2} - 3x - 54\)

\(\displaystyle P (9) = 9^{4} -6 \cdot 9 ^{3} + 2 \cdot 9 ^{2} - 3 \cdot 9 - 54\)

\(\displaystyle = 6,561 -6 \cdot 729 + 2 \cdot 81 - 3 \cdot 9 - 54\)

\(\displaystyle = 6,561 - 4,374 + 162 - 27- 54\)

\(\displaystyle = 2,268\)

\(\displaystyle P (9) \ne 0\), so \(\displaystyle x - 9\) is not a factor. 

By the Remainder Theorem, if a polynomial \(\displaystyle P(x)\) is divided by a binomial \(\displaystyle x-c\), the remainder is \(\displaystyle P(c)\). 

Let \(\displaystyle P(x)= x^{5} + x^{4} - x^{3} - x^{2}\). Setting \(\displaystyle c = - 2\) (since \(\displaystyle x + 2 = x - (-2)\) ), if \(\displaystyle P(x)\) is divided by \(\displaystyle x + 2\), the remainder is \(\displaystyle P(-2)\), which can be evaluated by setting \(\displaystyle x =- 2\) in the definition of \(\displaystyle P(x)\) and evaluating:

\(\displaystyle P(x)= x^{5} + x^{4} - x^{3} - x^{2}\)

\(\displaystyle P(2)= (-2)^{5} + (-2) ^{4} - (-2)^{3} - (-2)^{2}\)

\(\displaystyle =( -32 )+ 16 -( - 8) - 4\)

\(\displaystyle =-32 + 16+8- 4\)

\(\displaystyle = -12\)