P for Polynomial

Algebra Level 2

Given that $P(x)$ is a monic fifth-degree polynomial such that

$\begin{aligned} P(1) & = & 1^2 \\ P(2) & = & 2^2 \\ P(3) & = & 3^2 \\ P(4) & = & 4^2 \\ P(5) & = & 5^2,\end{aligned}$

find the value of $P(6)$ .

The answer is 156.

3 solutions

Eddie The Head
Dec 26, 2014

Clearly 1, 2, 3, 4, 5 are the roots of the polynomial $p(x) = x^2$ .

Because it is given that the leading co-efficient of $p(x)$ is 1, we must have that

$p(x) - x^{2} = (x-1)(x-2)(x-3)(x-4)(x-5)$ $p(x) = x^{2} + (x-1)(x-2)(x-3)(x-4)(x-5)$

$p(6) = 36+5*4*3*2*1$

Hence $p(6) = 156$ .

Having some doubts

$P(x) = x^2$ is not true for every $x \in R$
Why this information is given - leading coefficient 1?

U Z - 6 years, 5 months ago

Exactly! Thats why $P(x)-x^2$ is considered whose zeroes are $x=1,2,3,4,5$ . So $(x-1),(x-2),(x-3),(x-4),(x-5)$ must be the factor of $P(x)-x^2$ . Since it is of degree 5, no more $Q(x)$ is a factor.

Otherwise it could have been $P(x)-x^2=k(x-1)(x-2)(x-3)(x-4)(x-5)$ which also satisfies given condition. To let everyone know the value of $k$ , leading coefficient is given as 1

Pranjal Jain - 6 years, 5 months ago

Thanks understood

U Z - 6 years, 5 months ago

Your wording confused me for a while. You mean 1, 2, 3, 4, 5 are the solutions of the equation P(x)=x^2. Or, they are the roots of P(x)-x^2. Good solution.

Davy Ker - 5 years ago

I like this method, but here's mine, which amounts to the same thing, by way of a little algebra. This is based on the method of finite differences.

For an n'th degree polynomial, P(x), with leading coefficient a[n], when values of it are given for consecutive integers, the n'th differences are all just n!·a[n]. In this case, given that a[n] = 1 ("monic"), and n=5, this means that the fifth differences are all 5! = 120.

So, given 5 successive values, just make a table of differences down to the single 4th difference (which = 0), place a "120" below that on the 5th-difference line, and work back up by addition (brackets enclose these extended terms):

1 4 9 16 25 [156]

• 3 5 7 9 [131]

• • 2 2 2 [122]

• • • 0 0 [120]

• • • • 0 [120]

• • • • [120]

Of course, it follows by linearity of the method of finite differences, that this result must be just

6² + 5! = 36 + 120 = 156;

which is where it connects to Eddie The Head's solution.

Fred Shuman - 5 years ago

I think that first line was meant to be

Clearly 1, 2, 3, 4, 5 are the roots of the polynomial $p(x)-x^2$ .

David Lewis - 5 years, 1 month ago

sorry why cannot I use method of differences in this question?

ZHU YANKANG - 4 years, 10 months ago

Owen Scott
Mar 11, 2016

I see what you were going for with this problem, but I'm not sure that the answer has to be 156.

It takes 2 points to define a unique line (1st order polynomial). It takes 3 points to define a unique quadratic (2nd order polynomial). .... You need 6 points to define a unique 5th order polynomial.

Because only 5 points were provided, there are actually an infinite number of monic polynomials that can hit all 5 points provided. Thus, P(6) could have an infinite number of values.

Since it is given that the polynomial is monic, the leading coefficient has to be 1.Therefore, we have 5 degrees of freedom here and therefore need only 5 points to completely determine the polynomial.

Abdur Rehman Zahid - 5 years, 3 months ago

We know that the polynomial $P(x) - x^2$ has five zeroes: 1, 2, 3, 4, 5. This means that $P(x) = A(x)(x-1)(x-2)(x-3)(x-4)(x-5) + x^2,$ with $A(x)$ any polynomial. However, we know that the degree of $P(x)$ is five, so that the degree of $A(x)$ must be zero (i.e. it is a constant); we also know that $P(x)$ is monic, so that the coefficient of $x^5$ equals one; this implies that $A(x) = 1$ . Indeed, the solution is unique!

Arjen Vreugdenhil - 4 years, 9 months ago

Take line-case for instance, because we already knew y=x+m, so we just need 1 point to define this line. So.... Anyway, your critique is good. Thanks

Hoanthien Nguyen - 4 years, 7 months ago

Ani B
Jan 9, 2019

According to the given information, we know that $P(k)= k^2$ , for $k = 1, 2, 3, 4 ,5$ . Therefore, we know $f(k)-k^2 = 0$ , and has roots at $k = 1, 2, 3, 4 ,5$ . We can write that as a product of linear factors, and since it is monic, it has no coefficient multiplied by the product. Then, $f(k)-k^2 = (x-1)(x-2)(x-3)(x-4)(x-5)$ , and when plugging in the values for k, we get $5! + 36$ , which is equal to $156$ , and that is the answer.

P for Polynomial

The answer is 156.

3 solutions

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