Question

HELP ASAP : Find the largest value of n such that 5x^2+nx+48 can be factored as the product of two linear factors with integer coefficients

Answer 1

Answer:

$n=241$

Step-by-step explanation:

We are given

$5x^2+nx+48$

Let's assume it can be factored as

$5x^2+nx+48=(5x-s)(x-r)$

now, we can multiply right side

and then we can compare it

$5x^2+nx+48=5x^2-5rx-sx+rs$

$5x^2+nx+48=5x^2-(5r+s)x+rs$

now, we can compare coefficients

$rs=48$

$5r+s=-n$

$n=-(5r+s)$

now, we can find all possible factors of 48

and then we can assume possible prime factors of 48

$48=-+(1\times 48)$

$48=-+(2\times 24)$

$48=-+(3\times 16)$

$48=-+(4\times 12)$

$48=-+(6\times 8)$

Since, we have to find the largest value of n

So, we will get consider larger value of r because of 5r

and because n is negative of 5r+s

so, we will both n and r as negative

So, we can assume

r=-48 and s=-1

so, we get

$n=-(5\times -48-1)$

$n=241$