What polynomial function has a remainer of 4 when divided by (x+1) and a remainer of 3 when divided (x-2)?

Question

It wouuld be best if it was the smallest possible function

Answer 1

Try the lowest-order polynomial: Let P(x) = ax + b.

By the Remainder Theorem, the remainder when P(x) is divided by (x+1) is P(-1), which equals -a + b. This is to equal 4. So

P(-1) = -a + b = 4

The remainder when P(x) is divided by (x-2) is P(2), which is to equal 3. So

P(2) = 2a + b = 3

Solving this system gives a = -1/3, b = 11/3. So

P(x) = (-1/3) x + (11/3)

satisfies the conditions given.

If you want a polynomial whose leading coefficient is 1, you have to go to a higher-power polynomial.

So let Q(x) = x² + ax + b. We need Q(-1) = 4 and Q(2) = 3:

Q(-1) = 1 - a + b = 4
Q(2) = 4 + 2a + b = 3

That is,

-a + b = 3
2a + b = -1

The solution is a = -4/3, b = 5/3. So

Q(x) = x² - (4/3) x + (5/3)

also produces the desired remainders.

Answer 2

a million. (x^2-fifty 8)=(x^2-sixteen-40 two) =(x^2-sixteen)-40 two =(x^2-4^2)-40 two =(x-4)(x+4)-40 two which shows the rest is -40 two while (x^2-fifty 8)is split by utilising (x-4). 2. placed f(x)=0 =>(x-3)(x+a million)(x-5)=0 =>the two (x-3)=0, (x+a million)=0,(x-5)=0 =>x=3, x=-a million, x=5. so, zeros of the polynomial function are -a million,3,5