if 2x³+ax²+3x-5 and x³+x²-4x+9 leaves the same remainder deviding by x-2 find the value of a
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the answer A=3/4 i suppose
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The remainder of a polynomial can be found out by substituting the zero of the divider in the polynomial.
x³ + x² - 4x + 9 when divided by x - 2 leaves the remainder,
(2)³ + (2)² - 4(2) + 9 = 8 + 4 - 8 + 9
= 13.
Therefore, when 2 is substituted in 2x³ + ax² + 3x - 5, it should leave the remainder 13.
2(2)³ + a(2)² + 3(2) - 5 = 13
16 + 4a + 6 - 5 = 13
17 + 4a = 13
4a = 13 - 17
4a = -4
a = -1
∴ The required value of a is -1.
Hope my answer helps you..
x³ + x² - 4x + 9 when divided by x - 2 leaves the remainder,
(2)³ + (2)² - 4(2) + 9 = 8 + 4 - 8 + 9
= 13.
Therefore, when 2 is substituted in 2x³ + ax² + 3x - 5, it should leave the remainder 13.
2(2)³ + a(2)² + 3(2) - 5 = 13
16 + 4a + 6 - 5 = 13
17 + 4a = 13
4a = 13 - 17
4a = -4
a = -1
∴ The required value of a is -1.
Hope my answer helps you..
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