Polynomials - SAT Math

Notice that - (a-b) = b-a, so statement I & III are true after substituting the expression. Substitute the expression for statement II gives ((a+b)/(a-b))((a+b)/(b-a))=((a+b)(b+a))/((-1)(a-b)(a-b))=-1 〖(a+b)〗2/〖(a-b)〗2 =-((a+b)/(a-b))2 = -(a♦b)2 ≠ (a♦b)2

The best way to solve this problem is to plug in an appropriate value for a. For example, plug-in 11 for a because 11 divided by 7 will give us a remainder of 4.

Then 3_a + 5_, where a = 11, gives us 38. Then 38 divided by 3 gives a remainder of 2.

The algebra method is as follows:

a divided by 7 gives us some positive integer b, with a remainder of 4.

Thus,

a / 7 = b 4/7

a / 7 = (7_b +_ 4) / 7

a = (7_b_ + 4)

then 3_a + 5 =_ 3 (7_b_ + 4) + 5

(3_a_+5)/3 = \[3(7_b_ + 4) + 5\] / 3

= (7_b_ + 4) + 5/3

The first half of this expression (7_b_ + 4) is a positive integer, but the second half of this expression (5/3) gives us a remainder of 2.

Cancel by subtracting the exponents of like terms:

Set up an equation for the sentence: 15 – 3 = 2x and solve for x. X equals 6. If you plug in 6 for x in the expression 24/x, you get24/6 = 4. 4 is only choice greater than a.

Divide by by setting up a long division.

Divide the lead term of the dividend, , by that of the divisor, ; the result is

Enter that as the first term of the quotient. Multiply this by the divisor:

Subtract this from the dividend. This is shown in the figure below.

Repeat the process with the new difference:

Repeating:

The quotient - and the correct response - is .

It is not necessary to work a long division if you recognize as the sum of two perfect cube expressions:

A sum of cubes can be factored according to the pattern

,

so, setting ,

Therefore,

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.

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.

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:

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:

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.

First factor out a . Then the factors of the remaining polynomial,

, are and .

Set everything equal to zero and you get , , and because you cant forget to set equal to zero.

To find the degree of a polynomial we must find the largest exponent in the function.

The degree of the polynomial is 5, as the largest exponent of is 5 in the second term.

When a polynomial has more than one variable, we need to find the degree by adding the exponents of each variable in each term.

has a degree of 4 (since both exponents add up to 4), so the polynomial has a degree of 4 as this term has the highest degree.