Question

Iv) show that a = 7 and b = – 18, if y^3 + 10y^2 + ay + b leaves remainder zero if divided by y – 1 and y + 2.

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Answer 1

Answer:

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Answer 2

Answer:

Hi! Here is the step-by-step explanation!!

Step-by-step explanation:

$y^{3} + 10y^{2} + ay + b$ is exactly divisible by $y - 1$ as well as by $y + 2$. This means that :

$(1)^{3} + 10(1)^{2} + a(1) + b = 0 ------- (i)$

Also,

$(-2)^{3} + 10(-2)^{2} + a(-2) + b = 0 ------- (ii)$

From (i) and (ii), we get

$(1)^{3} + 10(1)^{2} + a(1) + b = (-2)^{3} + 10(-2)^{2} + a(-2) + b$

∴ $1 + 10 + a = -8 + 40 - 2a$

∴ $11 + a = 32 - 2a$

∴ $a + 2a = 32 - 11$

∴ $3a = 21$

∴ $a = 7$

Putting the value of a in equation (i), we get

$(1)^{3} + 10(1)^{2} + (7)(1) + b = 0$

∴ $1 + 10 + 7 + b = 0$

∴ $18 + b = 0$

∴ $b = -18$

Hence, proved

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