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If (3x + 1) 5 = ax5 + bx4 + cx3 + DX2 + ex + F, the value of A-B + C-D + E-F is __
Solution 1: (3x + 1) 5 = ax5 + bx4 + cx3 + DX2 + ex + F,
∴（3x+1）5=243x5+405x4+270x3+90x2+15x+1，
∴a-b+c-d+e-f=243-405+270-90+15-1=32．
Solution 2: let x = - 1, then (- 3 + 1) 5 = A-B + C-D + E-F = 32
So the answer is: 32
Given that the 5th power of (square of X - 3x + 1) = the 5th power of AX + the 4th power of BX + the 3rd power of Cx + the 2nd power of DX + ex + F, find the value of a + C + E
It is known that the 5th power of (square of X - 3x + 1) = the 5th power of AX + the 4th power of BX + the 3rd power of Cx + the 2nd power of DX + ex + F, let x = 1 (1-3 + 1) ^ 5 = a + B + C + D + e + F, so a + B + C + D + e + F = - 1 (1) let x = - 1 (1 + 3 + 1) ^ 5 = - A + B-C + D-E + F, so - a
Given the fourth power of AX + the third power of BX + the second power of Cx + DX + e = (X-2), find the value of a + B + C + D + E and a + C It's better to be more detailed. There are answers and processes for this on the Internet, but I don't understand. It's best to talk about how each formula value is calculated. More words are convenient for me to understand. For that (I don't quite understand how the first (X-2) ^ 4 = x ^ 4-8x ^ 3 + 24x ^ 2-32x + 16 changes. Why do you have to make x = 1? Can't it be equal to other numbers? I don't understand this.) I hope you can make it clear. I'll adopt what I understand. Forgive me for not being talented. So I want to understand every step, or I'll copy it directly.
From the binomial expansion theorem: (X-2) ^ 4 = x ^ 4-8x ^ 3 + 24x ^ 2-32x + 16
So: a = 1, B = - 8, C = 24, d = - 32, e = 16;
So: a + B + C + D + e = 1, a + C = 25;
The first finding a + B + C + D + e can directly make x = 1: a + B + C + D + e = (- 1) ^ 4 = 1;
If you don't understand, please hi me,
Known (3x-1) quartic = ax quartic + BX cubic + CX quadratic + DX + e, try to find the value of coefficient a + B + C + D + E
(3x-1)^4
=(9x^2-6x+1)^2
=81x^4+36x^2+1-108x^3+18x^2-12x
=81x^4-108x^3+54x^2-12x+1
Contrast
a=81 b=-108 c=54 d=-12 e=1
a+b+c+d+e
=81-108+54-12+1
=16
It is known that the quadratic + 3x + 1 of 3x cubic + ax can be divided by the square + 1 of X, and the quotient is 3x + 1. Then a =?
(3x+1)(x^2+1)
=3x^3+3x+x^2+1
=3x^3+x^2+3x+1
=3x^3+ax^2+3x+1
a=1
Find the quotient and remainder after dividing X-1 by the square + 5x + 3 of X to the 4th power - 3x to the 3rd power - 2x using the vertical formula Use the vertical!
The fourth power of X - the third power of 3x - the square of 2x + 5x + 3
=x^3(x-1)-2x^3-2x^2+5x+3
=x^3(x-1)-2x^2(x-1)-4x^2+5x+3
=x^3(x-1)-2x^2(x-1)-4x(x-1)+x+3
=x^3(x-1)-2x^2(x-1)-4x(x-1)+(x-1)+4
=(x^3-2x^2-4x+1)(x-1)+4
Quotient: x ^ 3-2x ^ 2-4x + 1
Remainder formula: 4
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