Question

Fin d the value of a when ay²-9y+4a is divided by 2y-1 gives the remainder 5upon 6

Answer 1

Given that (2y - 1) is the factor the given equation which leaves Remainder 5/6


By Remainder Theorem,

f(y) = f(1/2)

Hence,



$a {(y)}^{2} - 9y + 4a = \frac{5}{6} \\ \\ \\ a {( \frac{1}{2}) }^{2} - 9( \frac{1}{2} ) + 4a = \frac{5}{6} \\ \\ \\ \\ \frac{a}{4} - \frac{9}{2} + 4a = \frac{5}{6} \\ \\ \\ => \frac{a - 18 + 16a}{4} = \frac{5}{6} \\ \\ \\ => \frac{17a - 18}{4} = \frac{5}{6} \\ \\ \\ => 102a - 108 = 20 \\ \\ \\ \\ = > 102a = 20 + 108 \\ \\ => 102a = 128 \\ \\ \\ \\ => a = \frac{128}{102} \\ \\ \\ a = \frac{64}{51}$

Answer 2

Answer:

Step-by-step explanation:

P(y) = ay2-9y+4a

When P(y) is divided by 2y-1 it leaves a remainder 5/6, So

P(1/2) = 5/6 ( 2y-1=0, 2y=1, y=1/2 )

P(1/2) = 1/4a - 9/2 + 4a = 5/6

1/4a - 9/2 + 4a = 5/6

17/4a = 32/6 = 16/3

a = 16/3 * 4/17 = 64/51