Practice determining Factor Techniques

Which constant term would mean that the expression is completely factored?x2 - 3x + ____

10

What is the completely factored form of x^4y - 4x^2y - 5y? y(x^2 - 5)(x^2 + 1) y(x^2 + 5)(x^2 - 1) (x^2y - 5)(x^2 + 1) (x^2y + 5)(x^2 - 1)

A

Which polynomials are prime? Check all of the boxes that apply. x^2 + 9 x^2 - 9 x^2 + 3x + 9 -2x^2 + 8

A and C

Check all of the possible first steps in factoring a polynomial with four terms. factor out a GCF factor the difference of cubes factor a sum of cubes factor a difference of squares factor a perfect-square trinomial factor by grouping

A and F

A polynomial has two terms. Check all of the factoring methods that should be considered. common factor difference of cubes sum of cubes difference of squares perfect-square trinomial factoring by grouping

A, B, C, and D

A student factors a^6 - 64 to (a^2 - 4)(a^4 + 4a^2 + 16). Which statement about (a^2 − 4)(a^4 + 4a^2 + 16) is correct? The expression is equivalent and is completely factored. The expression is equivalent, but the (a^2 - 4) term is not completely factored. The expression is equivalent, but the (a^4 + 4a^2 + 16) term is not completely factored. The expression is not equivalent.

B

What is the completely factored form of 2x^2 - 32? (2x^2 + 16)(x - 16) 2(x + 4)(x - 4) 2(x + 8)(x - 4) 2(x - 8)(x - 4)

B

Which of the following statements about 42xy - 49x + 30y - 35 are true?Check all of the boxes that apply. One of the factors is (6y + 7). One of the factors is (7x + 5). One of the factors is (6y - 7). One of the factors is (7x - 5). This expression is prime.

B and C

What is the completely factored form of 25x^4 - 16y^2? (5x^4 + 4y)(5x - 4y) (5x^3 + 4y)(5x^2 - 4y) (5x^2 + 4y)(5x^2 - 4y) 25x^4 - 16y^2

C

Which expression does not factor? m^3 + 1 m^3 - 1 m^2 + 1 m^2 - 1

C

Describe the steps you would use to factor2x3 + 5x2 - 8x - 20 completely. Then state the factored form.

The polynomial may have solutions which are the divisors of -20, therefore -20 has the following divisors: .If x=1, then ,if x=-1, then ,if x=2, then , then x=2 is a solution and you have the first factor (x-2). If x=-2, then , then x=-2 is a solution, so you have the second factor (x+2).Since x-2 and x+2 are two factors of , then the polynomial is a divisor of and dividing the polynomial by you obtain .