
Solving cubic functions
Presentation
•
Mathematics
•
10th - 12th Grade
•
Medium
Standards-aligned
Jesus Molina
Used 24+ times
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6 Slides • 6 Questions
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By Jesus Molina
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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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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
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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
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Multiple Choice
What are the zeros of the function f(x) = -2x3 - 4x2 + 6x?
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Multiple Choice
3x^ 3 + 2x^ 2 − 12x − 8 = 0
{-2/3, -2, 2}
{-2, 2, 3}
{-2, -3}
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Multiple Choice
Solve the following equations:
x3+7x2−36=0
{2, -3, -6}
{2, 3, -6}
{-3, 2, 6}
None of the above
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Multiple Choice
Solve the following Equation:
2x3+5x2−14x−8=0
{2, -1/2, -4}
{-2, 1/2, 4}
{-3, 4, 6}
{3, -4, -6}
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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)
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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)
By Jesus Molina
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