Question

using factor theorem , show that (x-5) is the factor of the polynomial 2x³-5x²-28x+15​

Answer 1

Let, 2x^3-5x^2-28x+15= f (x)

where f (x) is the function of variable x.

If (x-5) is factor of f (x), let's consider that (x-5) is just equal to zero.

Therefore, x-5=0

x=5

Put value of x in f (x).

Therefore,

f (x) = [2. (5)^3] - [5. (5)^2] - [28. (5)] + 15

f (x) = 250 - 125 - 140 +15

f (x) = 0

As f (x) = 0, no one could ever stop me from writing that (x-5) is a factor of f (x) in this case.

Hope it helps you successfully...

Please mark as the brainliest...

Answer 2

Answer:

Step-by-step explanation:

factor theorem is used when factoring the polynomials completely. It is a theorem that links factors and zeros of the polynomial. According to factor theorem, if f(x) is a polynomial of degree n ≥ 1 and 'a' is any real number, then, (x-a) is a factor of f(x), if f(a)=0.

so finding f(5)

f(5)=2(5)³-5(5)²-28(5)+15

f(5)=2(125)-5(25)-28(5)+15

f(5)=250-125-130+15

f(5)=0

so, (x-5) is the factor of the polynomial 2x³-5x²-28x+15​