ANSWERS: 3

I MEANT (x1) NOT (x+1)

Synthetic division. I know you stated the question wrong, but suppose I do it for the question you asked, and then you can do it the same way for the correct question. ax^3  4x^2 + 5 x  3 is a polynomial It's a bit like a decimal number: 1248 is 1*10^3 + 2*10^2 + 4*10^1 + 8 If I divide it by 10 + 1, long hand: 1*10^3 + 2*10^2 + 4*10^1 + 8 1*10^3 + 1*10^2 (is 1*10^2 elevens) subtract: 1*10^2 + 4*10^1 + 8 1*10^2 + 1*10^1 (is 1*10^1 elevens) subtract: 3*10^1 + 8 3*10^1 + 3 (is 3 elevens) subtract: remainder = 5 Now for the polynomial divided by x+2: ax^3  4x^2 + 5 x  3 ax^3 + 2ax^2 (that's a*(x+2)) subtract: (4+2a)x^2 + 5x  3 (4+2a)x^2  (8+4a)x (that's ((4+2a))*(x+2)) subtract: (13+4a)x  3 (13+4a)x + (26+8a) (that's (13+4a)*(x+2)) subtract: 298a is the remainder Same again for x+1 ax^3  4x^2 + 5 x  3 ax^3 + ax^2 (that's a*(x+1)) subtract: (4+a)x^2 + 5x  3 (4+a)x^2  (4+a)x (that's ((4+a))*(x+1)) subtract: (9+a)x  3 (9+a)x + (9+a) (that's (9+a)*(x+1)) subtract: 22a is the remainder If I've not made any aritmetical mistakes, then I only need to set: 298a = 22a and solve to get the answer a = 1 Now you need to do this with the correct question.

Okay. There is something called a remainder theorem. Consider 2 equations ax^34x^2+5x3 and x+2. If we put x+2=0, we get x=2 ans if we put this x=2 in the equation ax^34x^2+5x3, we get the remainder when ax^34x^2+5x3 is divided by x+2. repeat this with x=1 for x+1 (as given in the question). we get 2 new equations, 8a16103 and a453. Since according to the question, both are equal, we have 8a16103 = a453. You can easily solve for a now.


1!Tellyx to be replaced with 1 instead of 1, (I saw the comment late) and proceed.

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