# Thread: Cubic polynomials- finding unknowns

1. ## Cubic polynomials- finding unknowns

I know how to utilise the remainder theorem and factor theorem on a basic level, but not sure how I'm suppose to approach this problem. Would appreciate if the method to solve this problem was shown to me step by step.

This problem should be simpler than the above problem.

2. A) If this cubic function has a turning point at (a, 0), that suggests to me that a is a repeating root and has an even multiplicity. (b, 0) is the other root, and since this is cubic, a has a multiplicity of 2, so this polynomial would be
$\displaystyle f(x) = d[(x - a)^2 (x - b)]$
(Normally, I would use the variable a in front instead of d, but a is already used as one of the roots.)

The y-intercept is $\displaystyle (0, -a^2 b)$, or $\displaystyle f(0) = -a^2 b$:
\displaystyle \begin{aligned} f(x) &= d[(x - a)^2 (x - b)] \\ -a^2 b &= d[(0 -a)^2 (0 - b)] \\ -a^2 b &= d(-a^2 b) \\ d &= 1 \end{aligned}
d = 1, so the leading coefficient of this cubic is 1:
$\displaystyle f(x) = (x - a)^2 (x - b)$

EDIT: Cleaned this up because I can't read.

01

3. $\displaystyle P(x) = x^3 + (2a - 3)x^2 - 2(3a - 1)x + 4a$

Use synthetic division to show that (x + 2a) is a factor:

Code:
-2a| 1   2a-3  -6a+2   4a
----    -2a     6a    -4a
----------------------
1     -3      2    0
We want to solve P(x) = 0. Well, $\displaystyle P(x) = (x + 2a)(x^2 - 3x + 2)$, so factor the quadratic:
\displaystyle \begin{aligned} 0 &= (x + 2a)(x^2 - 3x + 2) \\ &= (x + 2a)(x - 1)(x - 2) \end{aligned}

So the roots are x = -2a, x = 1, and x = 2.

01

4. B) The stationary point of inflection @ (-a, 0) suggests that the root x = -a has an odd multiplicity of 3 or more. The turning point @ (b, 0) suggests that x = b has an even multiplicity of 2 or more. We're looking for the lowest possible degree, so let's try a polynomial with a degree of 5:
$\displaystyle f(x) = d[(x + a)^3 (x - b)^2]$
(Okay, let's not multiply it out this time. )

The y-intercept is $\displaystyle (0, 2a^3 b^2)$, or $\displaystyle f(0) = 2a^3 b^2$:
\displaystyle \begin{aligned} f(0) &= d[(0 + a)^3 (0 - b)^2] \\ 2a^3 b^2 &= d(a^3 b^2) \\ d &= 2 \end{aligned}

So the polynomial is
$\displaystyle f(x) = 2[(x + a)^3 (x - b)^2]$.

01