Question

the sum and the product of the root of the equation are 0 and 1 are respectively 99x³=bx²+cx+d . find the value of d​

Answer 1

Step-by-step explanation:

If we divide a polynomial P(x) by another polynomial D(x) then we can find quotient Q(x) and remainder R(x) such that degree of R(x) is less than degree of divisor D(x) and P(x)=D(x)Q(x)+R(x).

Now, if we can find D(x) such that when D(x)=0, P(x) become 0 then the relationship shown above becomes 0=0 × Q(x)+R(x)=> 0=0+R(x)=>0=R(x).

Putting this in original relationship, we get P(x)=D(x)Q(x).

So, if D(x)=0=>P(x)=0 then D(x) is a factor of P(x).

It is given that P(x)=0 when x=-1; in other words P(x)=0 when (x+1)=0. Hence (x+1) is a factor.

Similarly, (x+3) and (x-5) are factors.

Ans.: (x+1), (x+3) and (x-5).

p.s.: Factorization is a(x+1)(x+3)(x-5)

Multiplying all 3 (except a), we get

a(x3+(1+3−5)x2+(3−15−5)x−15)

=a(x3−x2−17x−15)

Verification:

For x=-1, a(-1–1+17–15)=a x 0=0

For x=-3, a(-27–9+51–15)=a × 0=0

For x=5, a(125–25-85–15)=125 × 0=0