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