Polynomial 1,2,3.

Algebra Level 4

Let $f\left( x \right) ={ x }^{ 4 }+a{ x }^{ 3 }+b{ x }^{ 2 }+cx+d$ , for that $a,b,c,d$ are constants. If $f\left( 1 \right) =1993,\quad f(2)=3986,\quad f(3)=5979$ . Calculate the value of $\frac { 1 }{ 4 } [f(11)+f(-7)]$ .

The answer is 5233.

4 solutions

Chew-Seong Cheong
Dec 9, 2014

We note that:

$\begin {cases} f(1) = 1993 = 1\times 1933 \\ f(2) = 3986 = 2\times 1993 \\ f(3) = 5979 = 3\times 1993 \end {cases}$

So we can induce that:

$f(x) = (x-1)(x-2)(x-3)(x + \alpha) + 1933x$

$\Rightarrow \begin {cases} f(11) = (10)(9)(8)(11 + \alpha) + 1933(11) & = 29843 + 720\alpha \\ f(-7) = (-8)(-9)(-10)(-7 + \alpha) + 1933(-7) & = -8911 - 720\alpha \end {cases}$

$\Rightarrow \frac {1}{4} [f(11)+f(-7)] = \frac {1}{4} (29843 + 720\alpha -8911 - 720\alpha)$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \space = \frac {1}{4} (20932) = \boxed {5233}$

Nice solution!

Ellen Shang - 6 years, 6 months ago

Hi, can someone please explain where does the $(x-1)(x-2)(x-3)(x+\alpha)+1993x$ come from?

Pedro Varela - 6 years, 6 months ago

Since $f(x) = 1993x$ for $x = 1,2,3$ . We can think of $f(x)$ as $f(x) = (x-1)(x-2)(x-3)g(x) + 1993x$ . Notice that $(x-1)(x-2)(x-3)g(x) = 0$ , when $x = 1,2,3$ , therefore leaving $f(x) = 1993x$ , when $x = 1,2,3$ . Since $f(x)$ is a $4^{th}$ degree polynomial whose highest power of $x$ is $4$ , $g(x)$ must be of the $1^{st}$ degree, because $(x-1)(x-2)(x-3)$ is of the $3^{rd}$ degree. That is $g(x)$ is of the form $g(x) = \beta x + \alpha$ . And since $f(x)$ is a monolithic polynomial. that is the coefficient of $x$ of the highest power ( $x^4$ ) is $1$ . Therefore, $\beta$ must be $1$ , this implies $g(x) = x + \alpha$ .

$f(x) = (x-1)(x-2)(x-3)(x+\alpha) + 1993x$

$\quad \quad = x^4 + (1+2+3+\alpha)x^3 + (1\dot{}2+2\dot{}3+3\dot{}1+1\dot{}\alpha+2\dot{}\alpha+3\dot{}\alpha)x^2$ $\quad \quad \quad + (1\dot{}2\dot{}3 + 1\dot{}2\dot{}\alpha + 1\dot{}3\dot{}\alpha + 2\dot{}3\dot{}\alpha + 1993)x + 1\dot{}2\dot{}3\dot{}\alpha$

$\quad \quad = x^4 + (6+\alpha)x^3 + (11+6\alpha)x^2 + (1999 + 11\alpha )x + 6\alpha$

Please note that $f(x) = x^4 + (6+\alpha)x^3 + (11+6\alpha)x^2 + (1999 + 11\alpha )x + 6\alpha$ is of the form: $f(x) = x^4+ ax^3 + bx^2 + cx + d$ .

Pedro, you may want to practice more on polynomial on brilliant.org, such as: https://brilliant.org/practice/factor-polynomials-higher-degress/?subtopic=polynomials&chapter=polynomial-factoring

Chew-Seong Cheong - 6 years, 6 months ago

Cody Johnson
Dec 8, 2014

Let $g(x)=f(x)-x^4$ . Then $g(1)=1992$ , $g(2)=3970$ , and $g(3)=5898$ . Using an interpolating polynomial on this data set, we get $g(x)=-25x^2+2053x-36$ , so $f(x)=x^4-25x^2+2053x-36$ . Now compute $\frac{f(11)+f(-7)}4=\boxed{5233}$ .

Nice solution! I did the same once I noticed that ${x}^{4}$ reduces the problem.

Steven Zheng - 6 years, 6 months ago

Aareyan Manzoor
Dec 13, 2014

insert 1,2,3 into the polynomial and we get $\begin {cases} a+b+c+d=1992\\ 8a+4b+2c+d=3970-----(1)\\ 27a+9b+3c+d=5898 \end {cases}$ subtract 1 from 2, and 2 from 3. we get $\begin {cases} 7a+3b+c=1978\\ 19a+5a+c=1928 \end {cases}$ again subtract to get $12a+2b =-50---(2)$ now expand $f(11)+f(-7)=14641+1331a+121b+11c+d+2401-343a+49b-7c+d$ $f(11)+f(-7)=17042+988a+170b+2(2c+d)$ from ---(1), $2c+a=3970-8a-4a$ , sustitute to get $f(11)+f(-7) = 17042 +988a+170b+7940-16a-8b$ $f(11)+f(-7) = 24982+81(12a+2b)$ substitute from---(2) $f(11)+f(-7) = 24982+81 \times -50$ $f(11)+f(-7)=20932$ now $\dfrac{1}{4} \times 20932 = \boxed{5233}$

Nikola Djuric
Dec 8, 2014

a+b+c+d=1992 (1) , 8a+4b+2c+d=3970 (2) , 27a+9a+3c+d=5898 (3) 1/4(f(11)+f(-7))=1/4(14641+1331a+121b+11c+d+2401-343a+49b-7c+d)= 1/4(17042+988a+170b+4c+2d)= 1/2(8521+494a+85b+2c+d)

Now we want to try to express 494a+85b+2c+d (4) as linear combination of left sides of equalities (1),(2) and (3). (4)=x(1)+y(2)+z(3) which means 27x+8y+z=494 , 9x+4y+z=85 , 3x+2y+z=2 , x+y+z=1 System become x+y+z=1 , 2x+y=1 ,8x+3y=84 , 26x+7y=0 And finally x+y+z=1 , 2x+y=1 , 2x=81 , 12x=486 That means x=40.5 , y=-80 and z=40.5 ,so we can express (4) as combination (1),(2),(3). (4)=40.5(1)-80(2)+40.5(3) so we get
1/2(8521+494a+85b+c)=1/2(8521+40.5(5898+1992)-81x3970) =
1/2(8521+40.5x7890-81x3970)= 1/2(8521+81x3945-317600)=
1/2(8521+319545-317600)= 1/2(8521+1945) =10466/2 =5233 Answer is 5233.

Polynomial 1,2,3.

The answer is 5233.

4 solutions

0 pending reports