The polynomial P=mx^3+x^2+n is divided by the polynomial Q=x^2+x. Determine P?

Expert Answers
hala718 eNotes educator| Certified Educator

P= mx^3+x^2 +n

Q = x^2 +x

Since P is divided by Q, then Q is a factor for P, that is, Q's roots verifies P. Or if Q(r)=0, then P(r)=0

Q(r)= 0

==> r^2 +r = 0

==> r(r+1)=0

then r = {0,-1}

then 0 and -1 are roots for P

the P(0)=0

and P(-1)= 0

p(0) = N = 0

p(-1) = M(-1)^3 + (-1)^2 +0=0

p(-1) = -M +1 = 0

==> m= 1

Then P(x) = x^3 +x^2

 

neela | Student

P(x) = mx^3+x^2+n is divisible by x^2+x or x(x+1).

Therefore p(x) is divisible by x  Or x-0. So n = 0 ,as P(0)= 0= m*o^3+0^2+n.

P(x) is divisibe by x+1. Oe p(-1) = 0.Therefore,

P(-1) = 0 = m(-1)^3+(-1)^2+0. Or

-m+1 = 0. Or

m = -1.

Therefore P(x) = x^3+x^2

giorgiana1976 | Student

We'll write the rule of division with reminder.

P=Q*C+R

If the polynomial P is divided by Q, that means that the roots of Q are verifying the polynomial P, so that:

P(x1)=0 and P(x2)=0.

We'll find the roots of Q=x^2+x

x^2+x=0

We'll factorize:

x(x+1)=0

x1=0

x2=-1

P(0)=m*0+0+n

n=0

P(-1)=0

P(-1)=-m+1+n

-m+n+1=0, where n=0.

-m+1=0

-m=-1

m=1 and n=0

P=x^3+x^2