# Finding a,b,c,d in a quartic expression

Let $p(x)=x^4+ax^3+bx^2+cx+d$ where a,b,c,d are constants. If $p(1)=10$, $p(2)=20$, $p(3)=30$, compute $\frac {p(12)+p(-8)}{10}$. I have tried so far. \begin{align} a+b+c+d=&9\\8a+4b+2c+d=&4\\27a+9b+3c+d=&-51 \end{align} Manipulating these, I got $6a+b=-25$. Now, $$\frac {p(12)+p(-8)}{10}=\frac{24832+1216a+208b+4c+2d}{10}$$ $$=\frac{24832+202(6a+b)+(4a+4b+4c)+2b+2d}{10}$$ $$=\frac{19782+(36-4d)+2b+2d}{10}$$ $$=\frac{19818+2b-2d}{10}$$ How do I get rid of the $2b-2d$?

• Incidentally, the term for a degree $4$ polynomial is "quartic." Nov 1, 2013 at 14:23
• You made a slight error. You should have $2d$ in the numerator, and not $d$. Nov 1, 2013 at 14:29
• And I'm pretty sure that a biquadratic polynomial means a composition of two quadratic polynomials, that is $p(x)=q_1(q_2(x))$ where $q_1,q_2$ are quadratic. Nov 20, 2013 at 18:12

(2012's answer is correct, but the algebraic manipulations doesn't reveal what is happening. This answer explains why we could calculate the expression, despite not having enough information.)

By the remainder factor theorem, since $p(x) - 10x$ is a monic quartic polynomial with roots 1, 2, and 3, hence

$$p(x) - 10x = ( x-1) (x-2) ( x-3) (x-k),$$

where $k$ is some constant.

Hence, $p(12) - 120 = 11 \times 10 \times 9 \times (12-k)$ and $p(-8) - (-80) = (-9) \times (-10) \times (-11) \times (-8-k)$.

Note that the coefficients of $(12-k)$ and $-(-8-k)$ are the same, namely $11\times 10 \times 9$, so we can add them up to get:

$$p(12) + p(-8) = 120 + (-80) + 11 \times 10 \times 9 \times (12+8) = 19840.$$

If you want a similar problem to practice what you learnt in this problem, try this math problem on Brilliant.

As you noticed, you have three equations for four unknowns (a,b,c,d). Eliminate a, b and c expressing then as functions of d. Then compute your expression. By magics, d disappears ! Does this help ?

You have $p(x) = x^4 + ax^3 + bx^2 + cx + d$, and you're given that $a + b + c + d = 9$, $8a + 4b + 2c + d = 4$, and $27a + 9b + 3c + d = -51$

Now, $p(12) + p(-8) = 12^4 + 8^4 + (12^3 - 8^3) a + (12^2 + 8^2) b + (12 - 8) c + 2d = 24832 + 1216 a + 208 b + 4c + 2d = 24832 + 1216 a + 208 b + 2 (2c+d)$. You note that $2c + d = 4 - 8a - 4b$, and substitute that into the equation to get $p(12) + p(-8) = 24832 + 1216 a + 208 b + 8 - 16a - 8b = 24840 + 1200 a + 200b = 24840 + 200(6a + b).$

Plug in the $6a + b = -25$ and you get $p(12) + p(-8) = 24840 - 5000 = 19840$. Divide it by $10$ and you get $\displaystyle \frac{p(12) + p(-8)}{10} = 1984$.

Remember, $d + d \ne d$.

• Doing things differently,I found 1984 ! Nov 1, 2013 at 14:57
• @ClaudeLeibovici - I just assumed what Tejas found (i.e. with $6a + b = -30$) was correct. Is that wrong? Nov 1, 2013 at 15:01
• I find that 6 a + b = -25. But I started the problem from scratch. Then, there is probably an error in Teja's calculations. Nov 1, 2013 at 15:16
• I just checked and you are indeed correct. $6a + b = -25$. I'll go ahead and fix that in my answer. Nov 1, 2013 at 15:18