All PSAT Math Resources
Example Questions
Example Question #2 : Factoring Polynomials
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 #254 : New Sat
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 #6 : How To Factor A Polynomial
How many of the following are prime factors of the polynomial ?
(A)
(B)
(C)
(D)
Three
None
One
Two
Four
One
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.
Example Question #11 : Algebra
Completely factor the following expression:
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
.
Example Question #7 : How To Factor A Polynomial
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
〖–9a〗2 + 26ab +〖3b〗2) / 5
–〖9a〗2 + 27ab +〖3b〗2) / 5
(–9a2 – 27ab +3b2) / 5
(–9a2 – 28ab –3b2) / 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 #11 : Variables
Give the degree of the polynomial
The polynomial has one term, so its degree is the sum of the exponents of the variables:
Example Question #1 : Polynomials
Give the degree of the polynomial
The degree of a polynomial in more than one variable is the greatest degree of any of the terms; the degree of a term is the sum of the exponents. The degrees of the terms in the given polynomial are:
The degree of the polynomial is the greatest of these degrees, 100.
Example Question #12 : Variables
Give the degree of the polynomial
The degree of a polynomial in one variable is the greatest exponent of any of the powers of the variable. The terms have as their exponents, in order, 44, 20, 10, and 100; the greatest of these is 100, which is the degree.
Example Question #1 : Polynomial Operations
Give the degree of the polynomial
The degree of a polynomial in one variable is the greatest exponent of any of the powers of the variable. The terms have as their exponents, in order, 10, 20, 30, and 40; 40 is the greatest of them and is the degree of the polynomial.
Example Question #17 : Algebra
Which of these polynomials has the greatest degree?
All of the polynomials given in the other responses have the same degree.
All of the polynomials given in the other responses have the same degree.
The degree of a polynomial is the highest degree of any term; the degree of a term is the exponent of its variable or the sum of the exponents of its variables, with unwritten exponents being equal to 1. For each term in a polynomial, write the exponent or add the exponents; the greatest number is its degree. We do this with all four choices:
:
A constant term has degree 0.
The degree of this polynomial is 5.
The degree of this polynomial is 5.
The degree of this polynomial is 5.
The degree of this polynomial is 5.
All four polynomials have the same degree.