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Example Questions
Example Question #2171 : Sat Mathematics
Let and be integers, such that . If and , then what is ?
Cannot be determined
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.
Example Question #2172 : Sat Mathematics
How many negative solutions are there to the equation below?
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.
Example Question #1 : Factoring Polynomials
2x + 3y = 5a + 2b (1)
3x + 2y = 4a – b (2)
Express x2 – y2 in terms of a and b
〖–9a〗2 + 26ab +〖3b〗2) / 5
(–9a2 – 27ab +3b2) / 5
(–9a2 – 28ab –3b2) / 5
–〖9a〗2 + 26ab –〖3b〗2) / 5
–〖9a〗2 + 27ab +〖3b〗2) / 5
(–9a2 – 28ab –3b2) / 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
Example Question #31 : Polynomials
If the polynomial
is divided by
,
what is the remainder?
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:
Example Question #5 : Factoring Polynomials
Which of the following is a factor of the polynomial ?
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.
Example Question #7 : How To Factor A Polynomial
If the polynomial
is divided by
,
what is the remainder?
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:
Example Question #2173 : Sat Mathematics
Factor the following variable
(x2 + 18x + 72)
(x + 18) (x + 72)
(x – 6) (x + 12)
(x + 6) (x + 12)
(x + 6) (x – 12)
(x – 6) (x – 12)
(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)
Example Question #1 : Variables
Factor 9x2 + 12x + 4.
(9x + 4)(9x – 4)
(3x + 2)(3x – 2)
(9x + 4)(9x + 4)
(3x – 2)(3x – 2)
(3x + 2)(3x + 2)
(3x + 2)(3x + 2)
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 9x2 + 12x + 4 = 9x2 + 6x + 6x + 4
Let's look at the first two terms and last two terms separately to begin with. 9x2 + 6x can be simplified to 3x(3x + 2) and 6x + 4 can be simplified into 2(3x + 2). Putting these together gets us
9x2 + 12x + 4
= 9x2 + 6x + 6x + 4
= 3x(3x + 2) + 2(3x + 2)
= (3x + 2)(3x + 2)
This is as far as we can factor.
Example Question #2174 : Sat Mathematics
If , and , what is the value of ?
8
6
–6
0
–8
8
The numerator on the left can be factored so the expression becomes , which can be simplified to
Then you can solve for by adding 3 to both sides of the equation, so
Example Question #2175 : Sat Mathematics
Solve for x:
First, factor.
Set each factor equal to 0
Therefore,