Introduction of cubic function:
A Cubic function is a small different from a quadratic function. A Cubic functions have a 3 x intercept, The cubic function refer to as 3 degrees. The example of a cubic function is y=(x-1)(x+3)(x-4). it has 3 x intercepts which loaded on (1,0)(-3,0)(4,0).A cubic function is one of the functions which is formed as, F(x) =ax3+bx2+cx+dwhere a- nonzero (or) say polinomial of degree three. Quadratic function is derivation for cubic function. Also, a intergral for a cubic function.By ƒ(x) = 0 and assuming a ≠ 0 gives the cubic formula of the form:ax3+bx2+cx+d=0Coefficient a, b, c, d are real numbers. However, most of theory is also legal if they belong to field of characteristic other than 2 or 3.
Example Problems on Cubic Function:
Roots of a cubic function:
Each cubic equation with real coefficients have at least one solution x among the real numbers; this is a consequence of the Intermediate value theorem. We are able to differentiate several likely cases using the discriminant.
`Delta=18abcd-4b^3d+b^2c^2-4ac^3-36a^2d^2`
The next cases require to be measured
If Δ > 0, the equation have three distinct real roots.
If Δ = 0, the equation has a multiple root along with all its roots are real.
If Δ < 0, t the equation have one real root along with two non real complex conjugate roots.
Let us see some examples of cubic function:
Example 1:
Solving the factors of the cubic of the equation x3-3x2-25x+75.
Solution:
The given equation is x3-3x2-25x+75.
This form as ax3 + bx2 + cx + d
So, (x3-3x2) + (-25x+75)
Take a common variable:
=x2(x-3) -25(x-3)
=( x2-25) (x-3)
Here x2 – 25 in the form of a2 + b2 = (a + b) (a - b)
So, x2 – 25 = (x + 5)(x - 5)
=(x-5)(x+5)(x-3)
Answer: The solutions are 5,-5,3
Example 2:
Solving the factors of the cubic of the equation 6x3-36x2 = -54x
Solution:
This equation can be written as 6x3-36x2 + 54x=0.
This form as ax3 + bx2 + cx + d
So, 6x(x2-6x+9) =0.
Here x2 – 6x + 9 in the form of Ax2 + Bx + C so we find the factor.I like to share this Free math problem solver with you all through my article.
6x(x-3)(x-3) =0.
x=0, x=3, x=3.
Answer: The solutions are 0, 3, and 3.
Example 3:
Solving the factors of the cubic of the equation x3-4x2-100x+400.
Solution:
The given cubic equation is (x3-4x2) + (-100x+400)
=x2(x-4) -100(x-4)
=( x2-100) (x-4)
=(x-10)(x+5)(x-3)
Answer : The solutins are 5,-5,3
A Cubic function is a small different from a quadratic function. A Cubic functions have a 3 x intercept, The cubic function refer to as 3 degrees. The example of a cubic function is y=(x-1)(x+3)(x-4). it has 3 x intercepts which loaded on (1,0)(-3,0)(4,0).A cubic function is one of the functions which is formed as, F(x) =ax3+bx2+cx+dwhere a- nonzero (or) say polinomial of degree three. Quadratic function is derivation for cubic function. Also, a intergral for a cubic function.By ƒ(x) = 0 and assuming a ≠ 0 gives the cubic formula of the form:ax3+bx2+cx+d=0Coefficient a, b, c, d are real numbers. However, most of theory is also legal if they belong to field of characteristic other than 2 or 3.
Example Problems on Cubic Function:
Roots of a cubic function:
Each cubic equation with real coefficients have at least one solution x among the real numbers; this is a consequence of the Intermediate value theorem. We are able to differentiate several likely cases using the discriminant.
`Delta=18abcd-4b^3d+b^2c^2-4ac^3-36a^2d^2`
The next cases require to be measured
If Δ > 0, the equation have three distinct real roots.
If Δ = 0, the equation has a multiple root along with all its roots are real.
If Δ < 0, t the equation have one real root along with two non real complex conjugate roots.
Let us see some examples of cubic function:
Example 1:
Solving the factors of the cubic of the equation x3-3x2-25x+75.
Solution:
The given equation is x3-3x2-25x+75.
This form as ax3 + bx2 + cx + d
So, (x3-3x2) + (-25x+75)
Take a common variable:
=x2(x-3) -25(x-3)
=( x2-25) (x-3)
Here x2 – 25 in the form of a2 + b2 = (a + b) (a - b)
So, x2 – 25 = (x + 5)(x - 5)
=(x-5)(x+5)(x-3)
Answer: The solutions are 5,-5,3
Example 2:
Solving the factors of the cubic of the equation 6x3-36x2 = -54x
Solution:
This equation can be written as 6x3-36x2 + 54x=0.
This form as ax3 + bx2 + cx + d
So, 6x(x2-6x+9) =0.
Here x2 – 6x + 9 in the form of Ax2 + Bx + C so we find the factor.I like to share this Free math problem solver with you all through my article.
6x(x-3)(x-3) =0.
x=0, x=3, x=3.
Answer: The solutions are 0, 3, and 3.
Example 3:
Solving the factors of the cubic of the equation x3-4x2-100x+400.
Solution:
The given cubic equation is (x3-4x2) + (-100x+400)
=x2(x-4) -100(x-4)
=( x2-100) (x-4)
=(x-10)(x+5)(x-3)
Answer : The solutins are 5,-5,3
