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Given that the expression x³_ ax² + bx+c leaves the same remainder when divided by x + 1 or x-2, find 'a' in terms of 'b'.​

Sagot :

caylus

Answer:

Hi,

a=b+3

Step-by-step explanation:

[tex]\begin{array}{c||c|c|c|c}&x^3&x^2&x&1\\---&---&---&---&---\\&1&-a&b&c\\x=-1&&-1&a+1&-a-b-1\\---&---&---&---&---\\&1&-a-1&a+b+1&c-a-b-1\\\end{array}[/tex]

[tex]\begin{array}{c||c|c|c|c}&x^3&x^2&x&1\\---&---&---&---&---\\&1&-a&b&c\\x=2&&2&4-2a&2b+8-4a\\---&---&---&---&---\\&1&2-a&b+4-2a&c+2b+8-4a\\\end{array}[/tex]

Thus:

c+2b+8-4a=c-a-b-1

3b-3a=-9

a=b+3

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