One missing root

Algebra Level 2

Given the polynomial $P(x) = x^4 + mx^3 - 7x^2 + nx + p$ for constants $m,n,p$ such that $(x-1),(x-2),(x-3)$ divides $P(x)$ .

What is the value of $m+n+p$ ?

The answer is 6.

2 solutions

Janaki S
Apr 20, 2015

As (x-1) is a factor P(x), P(1) = 0

i.e., 1+m-7+n+p = 0

so, m+n+p = 6

Moderator note:

Nicely done. It will be quite handy to know that if $P(x)$ is a polynomial, then $P(1)$ is the sum of its coefficient.

How would you solve it if we replace $P(x)$ as $x^4 - mx^3 -7x^2 +nx + p$ ? Or reverse any of the signs of the constants?

Changing the signs of constants will still give me m+n+p using P(1) = 0

If the signs of non-constant coefficients are changed, then use the three factors given to get three equations.

eg: Using P(1) = 0 => -m+n+p = 6

    Using P(2) = 0 => -8m+2n+p = 12

     Using P(3) = 0 => -27m + 3n +p = -18

  Then, solve the above three equations to get m, n and p

Janaki S - 6 years, 1 month ago

I had it right in front of me, but I still formed a system of equations between all three solutions of x given :P. It worked, but it took needlessly longer lol.

Oli Hohman - 6 years, 1 month ago

Roger Erisman
Apr 22, 2015

Synthetic division using 1 leaves remainder of m + n + p - 6 = 0.

Therefore, value = 6.

One missing root

The answer is 6.

2 solutions

Moderator note:

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