WMC 2018 Senior Problem 32

Algebra Level 3

Let $f\left(x\right)=x^4+ax^3+bx^2+cx+d$ , where $f\left(1\right)=f\left(3\right)=f\left(-2\right)=1$ . Also, $f\left(-1\right)=0$ . Find $f(0)$ .

\frac{21}{4}

9

\frac{31}{4}

-\frac{1}{4}

\frac{25}{4}

2 solutions

Otto Bretscher
Oct 18, 2018

$f(1)=f(3)=f(-2)=1$ tells us that $f(x)=(x-1)(x-3)(x+2)(x-a)+1$ for some $a$ . Now $f(-1) =0$ gives $a=-\frac{7}{8}$ so $f(0)=\boxed{\frac{25}{4}}$ .

Jordan Cahn
Oct 17, 2018

Plugging in each given value yields an equation in terms of $a, b, c, d$ : $\begin{aligned} 1 + a + b + c + d &= 1 \\ 81 + 27a + 9b + 3c + d &= 1 \\ 16 - 8a + 4b - 2c + d &= 1 \\ 1 - a + b - c + d &= 0 \end{aligned}$ Putting these equations into a matrix yields $\begin{pmatrix} 1 & 1 & 1 & 1 & 0 \\ 27 & 9 & 3 & 1 & -80 \\ -8 & 4 & -2 & 1 & -15 \\ -1 & 1 & -1 & 1 & -1 \end{pmatrix} \xrightarrow{\text{RREF}} \begin{pmatrix} 1 & 0 & 0 & 0 & -\frac{9}{8} \\ 0 & 1 & 0 & 0 & -\frac{27}{4} \\ 0 & 0 & 1 & 0 & \frac{13}{8} \\ 0 & 0 & 0 & 1 & \frac{25}{4} \end{pmatrix}$ So $f(0) = d = \boxed{\frac{25}{4}}$ .

I find it easier if you do this:

Let $g(x) = f(x) - 1$ , then $g(x)$ has roots $1,3,-2$ . So, $g(x) = (x-k)(x-1)(x-3)(x+2)$ for some constant $k$ . Substitute $0-1 = f(-1) - 1 = g(-1) -1 = \cdots \Rightarrow k = -\frac78$ .

Pi Han Goh - 2 years, 7 months ago

Agreed, this is the intended solution.

Tiger Ang - 2 years, 7 months ago

WMC 2018 Senior Problem 32

2 solutions

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