All PSAT Math Resources
Example Question #1 : Factoring Polynomials
What is a possible value for x in x2 – 12x + 36 = 0 ?
There is not enough information
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
Example Question #2 : Factoring Polynomials
If r and t are constants and x2 +rx +6=(x+2)(x+t), what is the value of r?
It cannot be determined from the given information.
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.
Example Question #3 : Factoring Polynomials
Solve for :
Solving for simplifies to .
Because there are two 's, there needs to be two solutions, which is why we get .
Example Question #4 : 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.
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 #5 : Factoring Polynomials
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 #11 : Variables
How many of the following are prime factors of the polynomial ?
can be seen to fit the pattern
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 #1 : How To Factor A Polynomial
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 #6 : 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
–〖9a〗2 + 26ab –〖3b〗2) / 5
(–9a2 – 28ab –3b2) / 5
(–9a2 – 27ab +3b2) / 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