Factoring Trinomials, a is not 1 - Concept

Concept Concept (1)

Long division can be used to divide a polynomial by another polynomial, in this case a binomial of lower degree. When dividing polynomials, we set up the problem the same way as any long division problem, but are careful of terms with zero coefficients. For example, in the polynomial x^3 + 3x + 1, x^2 has a coefficient of zero and needs to be included as x^3+ 0x^2+3x+1in the division problem.

Sample Sample Problems (11)

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Factoring Trinomials, a is not 1 - Problem 1

Factor:

3x² − 2x − 8
Problem 1
How to factor trinomials when the leading coefficient has only one pair of factors.
Factoring Trinomials, a is not 1 - Problem 2

Factor:

6x² + 17x + 5
Problem 2
How to factor trinomials with the leading coefficient has more than one pair of factors.
Factoring Trinomials, a is not 1 - Problem 3

Factor:

-3x² − 17x − 10
Problem 3
How to factor trinomials when the leading coefficient is negative.
Factoring Trinomials, a is not 1 - Problem 4

Factor:

20x² + 80x + 35
Problem 4
How to factor trinomials when a monomial can be factored out first.
Factoring Trinomials, a is not 1 - Problem 5
Problem 5
Factoring with an area, or rectangle method when "a" is not one
Factoring Trinomials, a is not 1 - Problem 6
Problem 6
A method for factoring trinomials that always works, even if "a" is not one: using a diamond, and then factoring by grouping
Factoring Trinomials, a is not 1 - Problem 7
Problem 7
Factoring trinomials where the "a" value is prime, using a guess and method
Factoring Trinomials, a is not 1 - Problem 8
Problem 8
Guess and check and FOIL method for factoring trinomials where the "a" value is not prime
Factoring Trinomials, a is not 1 - Problem 9
Problem 9
Factoring trinomials with a greatest common factor the results in "a" equaling one
Factoring Trinomials, a is not 1 - Problem 10
Problem 10
Factoring trinomials with GCF and "a" is not 1
Factoring Trinomials, a is not 1 - Problem 11
Problem 11
A geometric interpretation of factoring trinoimals that uses a length times width equals area rectangular model. A "diamond puzzle" is used to find the rectangle's sub-areas.