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Question: Can someONE hELP ME ON A Couple of math questions.Factoring Polynomials?

Divide.
(3a^3 + 6a^2) divided by 3a

3a3 + 6a2 – 3a

3a3 + 6a2 + 3a

a2 + 2a

a2 + 6a2

5. Determine whether the polynomial is a perfect square and if it is, factor it.
n2 – 6n + 9

not a perfect square

is a perfect square: (n + 3)2

is a perfect square: (n – 3)2

is a perfect square: (n – 6)2

6. Determine whether the polynomial is a difference of squares and if it is, factor it.
y2 – 25

not a difference of squares

is a difference of squares: (y – 5)2

is a difference of squares: (y + 5)(y – 5)

is a difference of squares: (y + 5)2
thanks you guys

Answer: Divide.
(3a^3 + 6a^2) divided by 3a
= 3a³/3a + 6a²/3a
= a² + 2a

5. Determine whether the polynomial is a perfect square and if it is, factor it.
n² – 6n + 9
middle term = – 6n
= 2 ×√n³ × √9
= ± 2 × n × 3 = ± 6n
as middle term is – 6n it is a perfect square.
= (n – 3)²

6. Determine whether the polynomial is a difference of squares and if it is, factor it.
y² – 25
= (y)² – (5)²
= (y + 5)(y – 5)
is a difference of squares: (y + 5)(y – 5)
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Math Fractions & Equations : Factoring Trinomials

Question: Help algebra/math polynomials by degree 10 points!!!?

Classify the polynomial by degree and by number of terms

14
2x+3
-3x^2+6x-2
x^3-5
1-x^4

Answer: 14
Degree zero
1 term
monomial

2x+3
degree 1
2terms
binomial

-3x^2+6x-2
degree 2
3 terms
trinomial

x^3-5
degree 3
2 terms
binomial

1-x^4
degree 4
2 terms
binomial

all binomial,monomial and trionomial are polynomials

Algebra Tests – YourTeacher.com – 1000+ Online Math Lessons

The following Polynomial Factorization formulae are useful when trying to factor any polynomials. In math, it is useful to know some important formulae to help you solve math problems. You should be familiar with the Polynomial Factorization formulae below. Although, when each Polynomial Factorization equation is common sense when you multiply them out, it is useful to be able to recognize the common standard Polynomial Factorization equations such as those below. 

a ( c+d ) = ac + ad

(a + b)(a – b) = a 2 – b 2

(a + b)(a + b) = (a + b) 2 = a 2 + 2ab + b 2

(a – b)(a – b) = (a – b) 2 = a 2 – 2ab + b 2

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• (x + a)(x + b) = x ² + (a + b)x + ab
• (ax + b)(cx + d) = acx ² + (ad + bc)x + bd
• (a + b)(a + b)(a + b) = (a + b) ³ = a ³ + 3a ² b + 3ab ² + b ³ 
• (a – b)(a – b)(a – b) = (a – b) ³ = a ³ – 3a ² b + 3ab ² – b ³

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• (a – b)(a ² + ab + b ²) = a ³ - b ³
• (a + b)(a ² – ab + b ²) = a ³ + b ³
• (a + b + c) ² = a ² + b ² + c ² + 2ab + 2ac + 2bc
• (a – b)(a ³ + a ² b + ab ² + b ³) = a ⁴ - b ⁴
• (a – b)(a ⁴ + a ³ b + a ² b ² + ab ³ + b ⁴) = a ⁵ - b ⁵