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If the polynomials 2x3+ax2+3x−5 and x3+x2−4x+a leave the same remainder when divided by x−2, find the value of a

Question

If the polynomials 2x3+ax2+3x52x^3 + ax^2 + 3x - 5 and x3+x24x+ax^3 + x^2 - 4x + a leave the same remainder when divided by x2x - 2, find the value of aa.

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Solution

Sure, let's solve this step by step.

Step 1: According to the Remainder Theorem, a polynomial f(x) will leave a remainder of f(c) when divided by (x-c). So, if both polynomials leave the same remainder when divided by (x-2), we can say that they are equal when x = 2.

Step 2: Substitute x = 2 into both polynomials:

2(2)^3 + a(2)^2 + 3(2) - 5 = (2)^3 + (2)^2 - 4(2) + a

Step 3: Simplify both sides:

16 + 4a + 6 - 5 = 8 + 4 - 8 + a

Step 4: Further simplify:

15 + 4a = 4 + a

Step 5: Solve for a:

4a - a = 4 - 15

3a = -11

a = -11/3

So, the value of a is -11/3.

This problem has been solved

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