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

Question

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." — Cameron Buie, Nov 01 '13 at 14:23
You made a slight error. You should have $2d$ in the numerator, and not $d$. — Cameron Buie, Nov 01 '13 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. — tomasz, Nov 20 '13 at 18:12

score 3 · Answer 1 · answered Nov 01 '13 at 15:36

(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.

score 2 · Answer 2 · answered Nov 01 '13 at 14:59

2

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 ?

answered Nov 01 '13 at 14:59

Claude Leibovici

260,315

2012ssohn · Accepted Answer · 2013-11-01T15:19:28.317

1

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, $d + d \ne d$.

edited Nov 01 '13 at 15:19

answered Nov 01 '13 at 14:29

2012ssohn

3,827

Doing things differently,I found 1984 ! – Claude Leibovici Nov 01 '13 at 14:57
@ClaudeLeibovici - I just assumed what Tejas found (i.e. with $6a + b = -30$) was correct. Is that wrong? – 2012ssohn Nov 01 '13 at 15:01
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I find that 6 a + b = -25. But I started the problem from scratch. Then, there is probably an error in Teja's calculations. – Claude Leibovici Nov 01 '13 at 15:16
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I just checked and you are indeed correct. $6a + b = -25$. I'll go ahead and fix that in my answer. – 2012ssohn Nov 01 '13 at 15:18

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

3 Answers3