Solving cubic functions

Solving cubic functions

By Jesus Molina

The amount of possible solutions to a polynomial will be based on the degree of the polynomial. Ex. X^3 will have three possible solutions.

Rule of thumb

Standard Form​:

​f(x) = ax³ + bx² + cx + d

Cubic Functions Standard form. ​

Some text here about the topic of discussion

​There are three ways to solve cubic equations: Two are kind of special cases and one is the standard way.

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​Two special cases

Common factor:

• In this type there would be no constant term.

• Example 1 Solve for x: x^3 + 5x^2 – 14x = 0

• Solution x(x^2 + 5x – 14) = 0

• x(x + 7)(x – 2) = 0

• x = 0, x = 2, x = –7

Second Special case

Grouping:​

With this type, we must have all four terms of the cubic expression. We then pair terms with a common factor and see if a common bracket emerges.

• ​​Solve for x: x^3 + 2x^2 – 9x – 18 = 0

• (x3 + 2x2 ) – (9x + 18) = 0​

• ​x^2 (x + 2) – 9(x + 2) = 0

• ​(x + 2)(x2 – 9) = 0

• ​(x + 2)(x – 3)(x + 3) = 0

• ​x = –2, x = 3, x = –3

Multiple Choice

What are the zeros of the function f(x) = -2x³ - 4x² + 6x?

• x = 0, -2, -3, 1
• x = -3, 1
• x = 0, 3, -1
• x = 0, -3, 1

Multiple Choice

3x^ 3 + 2x^ 2 − 12x − 8 = 0

• {-2/3, -2, 2}

• {-2, 2, 3}

• {-2, -3}

Multiple Choice

Solve the following equations:

$$x^3+7x^2-36=0$$  

• {2, -3, -6}

• {2, 3, -6}

• {-3, 2, 6}

• None of the above

Multiple Choice

Solve the following Equation:

$$2x^3+5x^2-14x-8=0$$  

• {2, -1/2, -4}

• {-2, 1/2, 4}

• {-3, 4, 6}

• {3, -4, -6}

Multiple Choice

Factor using Factoring by Grouping:

2n⁴ + 7n³ - 4n - 14

• (n - 2)(n - 2)(n - 2)(2n + 7)

• (n - 2)³(2n + 7)

• (n³ - 2)(2n + 7)

• (n + 1)(n - 1)(2n + 7)

Multiple Choice

Factor using Factoring like a Quadratic:

k⁴ - 10k² - 39

• (k² - 13)(k² + 3)

• (k - 13)(k + 3)

• (k² + 15)(k² - 3)

• (k + 13)(k - 3)