All Algebra II Resources
Example Questions
Example Question #141 : Polynomials
Factor .
In order to factor, we need to find two numbers that add up to , and multiply together to get . Because the signs of those numbers are both negative, one of the numbers we're looking for will be positive, and one will be negative. It helps if we first look at all of the pairs of the factors of :
We know that these pairs will multiply to get , we just need to narrow it down to the pair that, when added, will give . We find that the pair does that. We plug those numbers into our answer:
Example Question #141 : Polynomials
Factor .
The first thing we can do is factor a out of each term:
Now we need to find two numbers that add up to , and when multiplied make . We can look at the pairs of factors of :
Of these pairs, only one will equal when added together, so that's the one we're looking for. To finish, we plug those numbers into our function:
Example Question #141 : Polynomials
Factor the following equation using the AC method.
Using the AC method, A=8,B=-2 C=-6, therefore A*C=-48. So you must find 2 factors of -48 which multiply and add/subtract to get the B term which is -2. So going through the factors of -48 you come to the factors of -8 and +6, which multiply together to give you -48 and add together to give you -2.
Then you insert factors as follows:
Factor by grouping which is as follows:
Then factor
Then
Example Question #92 : Factoring Polynomials
Factor the polynomial
To solve this, first find what numbers give a product of 20 and when added yield -12.
Since the latter is negative, both numbers are negative.
and
Therefore
is the solution and can be verified by distributing:
Example Question #141 : Intermediate Single Variable Algebra
Factor the following polynomial
Expression cannot be factored
To simplify the factorization pull out an x term
Find two numbers that add to 16 and multiply to 60.
Example Question #146 : Polynomials
Factor .
To begin, let's clear up the from the quadratic term by multiplying the entire numerator and denominator by :
We can distribute the through the numerator:
Let's factor the out of the denominator now, just to clear the problem up a bit:
Now that we have the equation in a more recognizable form, we need to find two numbers that add up to , and multiply to get . Let's look at the factor pairs of :
We know that these pairs multiply to get (if we give one of the numbers a negative sign), so we need to find the pair that adds to . Because the three is negative, we know that the larger number will have the negative sign. We find that:
So our numbers are and . Now we can plug them back into our equation to get a final answer of:
Example Question #143 : Polynomials
Factor the polynomial
The polynomial will be easier to factor with an x pulled out
Two numbers are needed that add to 15 and multiply to 56. Trial and error is show that those two numbers are 7 and 8.
Example Question #148 : Polynomials
Factor:
To factor this polynomial, first take a common factor of negative one in order to change the coefficient of to positive.
Factor the terms inside the parentheses.
The possible factors of the last term, 42, that will provide the middle term are .
Factorize the polynomial.
Distribute the negative one through the first binomial.
The answer is:
Example Question #149 : Polynomials
Find the zeros of the function f(x) where...
The easiest way to solve this problem is to factor the original function, and then to find the zeroes from the factored form. To do this, we start with the original function, f(x).
Next, we need to set up the function in factored form, leaving blanks for the numbers we don't yet know.
At this point, we need to find two numbers - one for each blank. By looking at the original function, we can gather a few clues that will help us find the two numbers. The product of these two numbers will be equal to the last term of our original function (-14, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of our original function (-5, or b in the standard quadratic formula). Because their product is negative (-14), that must mean that they have different signs - otherwise, their product would be positive. Also, because their sum is negative (-5), we know that the larger of the two numbers must be negative, otherwise their sum would be positive.
Now, at this point, we may need to test a few different possibilities, using the clues we gathered from the original function. In the end, we'll find that the only numbers that work here are 2 and -7, as the product of 2 and -7 is -14, and sum of 2 and -7 is -5. So, this results in our function's factored form looking like...
Now, the final step in this problem is finding the zeros. To do this, we need to think about what a zero is. A zero is the x-value(s) at which...
So, to solve for our zeros, we just need to set the right side of our function equal to zero and solve for x.
Because if either of these two factors is equal to zero, the entire function will be equal to zero (as anything multiplied by zero is zero), we can consider each of the two factors separately and solve for x. We'll start with the factor on the left.
We'll finish with the factor from the right.
Now, we have both of our zeros and the answer to our problem...
Example Question #150 : Polynomials
Find the zeros of the function q(x) where...
None of the other answers
The easiest way to solve this problem is to factor the original function, and then to find the zeroes from the factored form. To do this, we start with the original function, f(x).
Next, we need to set up the function in factored form, leaving blanks for the numbers we don't yet know.
At this point, we need to find two numbers - one for each blank. By looking at the original function, we can gather a few clues that will help us find the two numbers. The product of these two numbers will be equal to the last term of our original function (24, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of our original function (11, or b in the standard quadratic formula). Because their product is positive (24), that must mean that they have the same signs - otherwise, their product would be negative. Also, because their sum is positive (11), and we already know their signs are the same because their product is positive, we know that their signs must both be positive, as the sum of two negative numbers is negative.
Now, at this point, we may need to test a few different possibilities using the clues we gathered from the original function to guide our testing. In the end, we'll find that the only numbers that work here are 3 and 8, as the product of 3 and 8 is 24, and sum of 3 and 8 is 11. So, this results in our function's factored form looking like...
Now, the final step in this problem is finding the zeros. To do this, we need to think about what a zero is. A zero is the x-value(s) at which...
So, to solve for our zeros, we just need to set the right side of our function equal to zero and solve for x.
Because if either of these two factors is equal to zero, the entire function will be equal to zero (as anything multiplied by zero is zero), we can consider each of the two factors separately and solve for x. We'll start with the factor on the left.
We'll finish with the factor from the right.
Now, we have both of our zeros and the answer to our problem...
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