Question

one of the zeroes of the polynomial x^2-mx+n is 4 such that n is a positive integer divisible by 7 and less than 50 and m is a prime number. If one of the zeroes of the polynomial x^2-mx+s is 9, then find the value of s. answer fast

Answer 1

Answer:

$s=-180$

Step-by-step explanation:

Since 4 is is a zero of $x^2+mx+n$, then $16+4m+n=0$.

Solving for m, $m=-\dfrac{n+16}{4}$.

Since n is positive, divisible by 7, and less than 50, then n=7,14,21,28,35,42,49.

The only value of n that will result in a prime value for m is n=28.

Which means that m=-11.

Now, $x^2+11x+s$ has 9 as a root,

this means that $81+99+s=0$.

Which results in $s=-180$.