Question

Factor the following polynomials completely. Show work below.

Answer 1

The factor form of the polynomials are: Case 1: (x - 10) · (x - 3), Case 2: (2 · x + 7) · (2 · x - 7), Case 3:
`3 \cdot (x - 3) \cdot \left(x + (2)/(3) \right)`, Case 4:
`4 \cdot (2 \cdot x + 5) \cdot \left(x - (9999)/(40) \right)\cdot \left(x - (1)/(40)\right)`, Case 5: 3 · x · (x + 10) · (x + 3), Case 6: (x² + 7) · (x + 2) · (x - 2)

How to factor polynomials

Herein we find six cases of polynomials, whose factor form must be derived according to following rules:

Common factor

a · x² + a · b = a · (x² + b)

Difference of squares

a² - b² = (a + b) · (a - b)

Sum of cubes

a³ + b³ = (a + b) · (a² - a · b + b²)

Factor of quadratic equations

a · x² - a · (r₁ + r₂) · x + r₁ · r₂ = a · (x - r₁) · (x - r₂)

Now we proceed to factor the polynomials:

Case 1: x² - 13 · x + 30

x² - 13 · x + 30

(x - 10) · (x - 3)

Case 2: 4 · x² - 49

4 · x² - 49

(2 · x + 7) · (2 · x - 7)

Case 3: 3 · x² - 7 · x - 6

3 · x² - 7 · x - 6

`3 \cdot \left(x^2 - (7)/(3)\cdot x - 2 \right)`

`3 \cdot (x - 3) \cdot \left(x + (2)/(3) \right)`

Case 4: 8 · x³ + 125

8 · x³ + 125

(2 · x + 5) · (4 · x² - 1000 · x² + 25)

`4 \cdot (2 \cdot x + 5) \cdot \left(x^2 - 250 \cdot x + (25)/(4)\right)`

`4 \cdot (2 \cdot x + 5) \cdot \left(x - (9999)/(40) \right)\cdot \left(x - (1)/(40)\right)`

Case 5: 3 · x³ + 39 · x² + 90 · x

3 · x³ + 39 · x² + 90 · x

x · (3 · x² + 39 · x + 90)

3 · x · (x² + 13 · x + 30)

3 · x · (x + 10) · (x + 3)

Case 6: x⁴ + 3 · x² - 28

x⁴ + 3 · x² - 28

(x² + 7) · (x² - 4)

(x² + 7) · (x + 2) · (x - 2)