Let the polynomial f (x) be divided by X-1 and divided by x-3, then find the remainder of F (x) divided by x ^ 2 + 2x-3 If you can, I hope it can include process, thx~

The polynomial f (x) is divided by X-1, so f (x) = (x-1) g (x) is divided by (x + 3) CO12, so f (- 3) = - 4G (- 3) = 12, that is g (- 3) = - 3, so g (x) = (x + 3) H (x) - 3f (x) = (x-1) (x + 3) H (x) - 3 (x-1) = (x ^ 2 + 2x-3) H (x) - 3 (x-1) f (x) is divided by x ^ 2 + 2x-3

The remainder of the polynomial f (x) divided by (x-1), (x + 1) and (x + 2) are - 1,1,2 respectively. Find the remainder of F (x) divided by (x-1) (x + 1) (x + 2)

The remainder of F (x) divided by (x-1) is - 1, or X
The remainder of F (x) divided by (x + 1) is 1, or X
The remainder of F (x) divided by (x + 2) is 2, or X
So: the remainder of F (x) divided by (x-1) (x + 1) (x + 2) is X

If the polynomial f (x) divided by (x-1) ^ 2 (x + 2) ^ 2 is 3x + 2 and 5x-3 respectively, then the remainder of F (x) divided by (x-1) ^ 2 (x + 2) is
I have seen a problem-solving step, that is, if f (1) = 5, X-1 can divide f (x) - 5, then there exists g (x) such that
F (x) - 5 = (x-1) g (x), f (x) - 5 divided by (x-1) & # 178; the remainder is 3x + 2-5 = 3 (x-1), then G (x) divided by X-1, the remainder is 3, G (1) = 3
I really don't understand why g (x) divided by X-1 is 3,

F (x) - 5 divided by (x-1) &# 178; the remainder is 3x + 2-5 = 3 (x-1);
So there exists H (x) such that f (x) - 5 = H (x) (x-1) &# 178; + 3 (x-1)
Because f (x) - 5 = (x-1) g (x), we know h (x) (x-1) &# 178; + 3 (x-1) = (x-1) g (x)
We can get g (x) = H (x) (x-1) + 3
So g (x) divided by X-1 is 3

It is known that the remainder of F (x) divided by (x-1) and (X-2) is 1 and 2 respectively. Try to find the remainder of F (x) divided by (x-1) (X-2)

It is known that the remainder of polynomial f (x) divided by (x-1) and (X-2) is 1 and 2 respectively. Try to find the remainder of polynomial f (x) divided by (x-1) (X-2)
[solution] it is not difficult to know that the remainder of F (x) divided by (x-1) (X-2) must be a linear expression, so it can be set as ax + B
f(x)＝(x－1)(x－2)g(x)+ax+b
According to the conditions of the title, when x = 1, f (x) = 1; when x = 2, f (x) = 2
a+b=1,2a+b=2
The solution is a = 1, B = 0, so the remainder is X

Find: the remainder of polynomial f (x) divided by X-1, X-2, x-3 is 1, 2, 3 respectively, then what is the remainder of polynomial f (x) divided by (x-1) (X-2) (x-3)?

The remainder of polynomial f (x) divided by X-1, X-2 and x-3 are 1, 2 and 3 respectively
f(1)=1,f(2)=2,f(3)=3
This shows that 1,2,3 are the three roots of the equation f (x) - x = 0
So f (x) - x must have the factor (x-1) (X-2) (x-3)
So what we're looking for is X

Given that the remainder of the polynomial f (x) divided by X + 2 is 1; the remainder of the polynomial f (x) divided by X + 3 is - 1, then the remainder of the polynomial f (x) divided by (x + 2) (x + 3) is?
Options: a.2x-5 b.2x + 5 c.x-1 D.X + 1 e.2x-1

Choose B
According to the first condition, f (x) = a (x + 2) + 1 = (A-1) (x + 2) + (x + 3)
According to the second condition, f (x) = B (x + 3) - 1 = (B-1) (x + 3) + (x + 2)
(where a and B are integers of x)
Then (A-1) (x + 2) + (x + 3) = (B-1) (x + 3) + (x + 2)
The results show that (A-2) (x + 2) = (b-2) (x + 3)
It is known that the integral A-2 contains the factor X + 3, and B-2 contains the factor X + 2
Then f (x) = (A-1) (x + 2) + (x + 3) = (A-2) (x + 2) + (2x + 5)
Since A-2 contains the factor X + 3, then (A-2) (x + 2) can be divisible by (x + 2) (x + 3),
So the remainder is 2x + 5

Given that the remainder of the polynomial f (x) divided by X + 2 is 1; the remainder of the polynomial f (x) divided by X + 3 is - 1, then the remainder of the polynomial f (x) divided by (x + 2) (x + 3) is 1
I found an answer on the Internet, but there is one place I can't understand
6 divided by 4 leaves 2
Can it be written as 6 = 4 * 1 + 2
Can 7 divide by 5 and 2 be written as 7 = 5 * 1 + 2
Can f (x) be divided by G (x) and a be written as f (x) = g (x) * r (x) + a
R (x) is an integral
If f (x) is divided by X + 2 and the remainder is 1, is it written as f (x) = (x + 2) * r (x) + 1
So if we take x = - 2 into f (- 2) = 1
If you divide X by X + 3, the remainder is - 1, which is equivalent to f (- 3) = - 1
According to the above conclusion, can I set my requirements for him as
f（x）=（x+2）（x+3）*G（x）+ax+b（!）
Ax + B is the one we want
Because his divisor is a polynomial with X
So you can't just set a to one dollar a time
Take x = - 2 and x = - 3 in
f（-3）=-1
f（-2）=1
－2a＋b＝1
－3a＋b＝－1
We can calculate a and B
It is: a = 2, B = 5
So it is: 2x + 5
This is where the exclamation mark is. Why can he be sure that the remainder is a one-time form? Why should he set the remainder as ax + B? I can see it all except here

In the division of numbers, the remainder must be smaller than the divisor, otherwise it is the remainder
In the division of polynomials, the degree of remainder must be lower than that of division
The division formula (x + 2) (x + 3) is quadratic, so the highest degree of the remainder formula should be once, and the remainder formula can also be constant

Polynomial f (x) divided by X-1, X-2, x-3, the remainder is 1,2,3 respectively. Find the remainder of F (x) divided by (x-1) (X-2) (x-3)

Let f (x) = 1 + a (x-1) + B (x-1) (X-2) + (x-1) (X-2) (x-3) g (x)
Then f (2) = 2, f (3) = 3, we can get
A = 1, B = 0, then
f(x)=x+(x-1)(x-2)(x-3)g(x)
That is to say, the remainder of F (x) divided by (x-1) (X-2) (x-3) is X

The remainder of polynomial f (x) divided by X + 1, X-2 are 2 and 5 respectively. Find the remainder of polynomial f (x) divided by (x + 1) (X-2)

If f (x) is divided by X + 1, the remainder of X-2 is 2 and 5 respectively, then f (x) - 2 = K1 (x + 1), (x + 1) = [f (x) - 2] / k1f (x) - 5 = K2 (X-2), (X-2) = [f (x) - 5] / K2 (x + 1) (X-2) = [f (x) - 2] / K1 * [f (x) - 5] / K2 = [f ^ 2 (x) - 7F (x) + 10] / (k1k2)

(0.11-x) * [(0.18-2x) square], and then divided by [(0.22-2x) square], finally equal to 0.81, thanks

（0.11-x）(0.18-2x)²/(0.22-2x)²=0.81
4(0.11-x)(0.09-x)²/[4(0.11-x)²]=0.81
(0.09-x)²/(0.11-x)=0.81
The reduction is: X & # 178; + 0.63x-0.081 = 0
Multiply both sides by 1000:1000x & # 178; + 630-81 = 0
(2x+27)(x-11)=0
So x = 27 / 2 or 11

Let the polynomial f (x) be divided by X-1 and divided by x-3, then find the remainder of F (x) divided by x ^ 2 + 2x-3 If you can, I hope it can include process, thx~