All Algebra 1 Resources
Example Questions
Example Question #5 : Solving By Factoring
Factor the following expression.
This expression involves the difference of two cubic terms. To factor an expression in this format, we can use a special formula.
Before we can use this formula, we need to manipulate our original expression to identify and .
Comparing this with the formula, and . Now we can use the formula to factor.
Example Question #341 : Variables
Factor completely:
can be factored out of each term, so do this first:
is the sum of squares and cannot be factored, so is the correct factorization.
Example Question #342 : Variables
Factor completely:
can be factored out of each term, so do this first:
The second factor can be factored as the difference of two squares:
Example Question #343 : Variables
Factor completely:
The polynomial is prime.
This is a difference of squares and can therefore be factored using the appropriate pattern:
Example Question #344 : Variables
Factor the following polynomial:
The coefficients are 20, 5, and 10. The greatest common factor of those three numbers is 5. Factor out the 5, making sure to keep the signs the same inside the parentheses:
Now factor out the term:
Example Question #345 : Variables
Factor by grouping.
Not factorable
Looking at the first two terms, the greatest common factor is 3b, hus we can factor out 3b.
Looking at the last two terms, the greatest common factor is –1. Factor out the –1.
Notice that you can factor out the .
This is your final answer.
Example Question #5 : Solving By Factoring
Factor the following expression.
Not factorable
This problem involves the difference of two cubic terms. We need to use a special factoring formula that will allow us to factor this equation.
But before we can use this formula, we need to manipulate to make it more similar to the left hand side of the special formula. We do this by making the coefficients (343 and 64) part of the cubic power.
Comparing this with , and .
Plug these into the formula.
Example Question #346 : Variables
Factor .
To factor this polynomial, first think of the pairs of numbers that when multiplied, equal .
and are a good place to start.
Then, you know that one of them has to be negative since your last term in the polynomial is negative. Try different combinations to get a . This would mean that you have a and .
Thus, your answer is .
Example Question #341 : Polynomials
Factor this polynomial:
First, a term can be factored out to give a simplified polynomial:
Then, the polynomial can be factored to give the correct answer:
Example Question #5 : Quadratic Equations
Factor:
Begin by factoring out a 2:
Then, we recognize that the trinomial can be factored into two terms, each beginning with :
Since the last term is negative, the signs of the two terms are going to be opposite (i.e. one positive and one negative):
Finally, we need two numbers whose product is negative thirty-five and whose sum is positive two. The numbers and fit this description. So, the factored trinomial is:
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