If y3 +ay2 by +6 is …
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Sia ? 5 years, 4 months ago
Let {tex}p\left( y \right){\text{ }} = {\text{ }}{y^3} + {\text{ }}a{y^2} + {\text{ }}by{\text{ }} + {\text{ }}6{/tex}
p(y) is divisible by y – 2
Then P (2) = 0
{tex}{2^3} + a \times {2^2} + b \times 2 + 6 = 0{/tex}
{tex}8 + 4a + 2b + 6 = 0{/tex}
{tex}{\text{4a + 2b = }} - {\text{14}}{/tex}
{tex}{\text{2a + b = }} - {\text{7 (i)}}{/tex}
If p (y) is divided by y-3 remainder is 3
{tex}\therefore {\text{ p (3) = 3}}{/tex}
{tex}{{\text{3}}^3} + a \times {3^2} + b \times 3 + 6 = 3{/tex}
9a+3b=-30
3a+b=-10 ---(ii)
Subtracting (i) from (ii)
-a = 3 and a = -3
Put a = -3 in eq (i)
{tex}2 \times - 3{\text{ }} + {\text{ }}b{\text{ }} = {\text{ }} - 7{/tex}
-6+b=-7
b=-7+6
b=-1
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