Question

find the value of k for which the cubic polynomial 3y^3 - 3/2y^2 + ky + 5 is exactly divisible by (y- 1/2)​

Answer 1

Answer:

k=-10

Step-by-step explanation:

y=½

3x(½)³-3/2x½²+½k+5 =0

⅜-⅜+k/2+5 =0

k/2+5 =0

k/2=-5

k=-10

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Answer 2

Given,

A polynomial 3y³ - 3/2y² + ky + 5 is exactly divisible by (y- 1/2)​.

To find,

The value of k

Solution,

Since, (y-1/2)=0 is the root of the polynomial 3y³ - 3/2y²+ ky + 5.

Hence, y = 1/2 would satisfy the given polynomial.

Now,

⇒   3y³ - 3/2y² + ky + 5 = 0

⇒   3(1/2)³ - 3/2(1/2)² + (1/2)k + 5 = 0

⇒   3(1/8) - 3 / 2(1/4) + (1/2)k + 5 = 0

⇒   3/8  - 6 + (1/2)k + 5 = 0

Taking the LCM of left hand side of the equation as 40

⇒   (3 - 48 + 4k + 40) / 8 = 0

⇒   4k - 5 = 0

-5 would shift from LHS to RHS and its sign would change from -5 to +5.

⇒   4k - 5 = 0

⇒   4k = 5

⇒   k = 5/4.

Hence, the value of k = 5/4 for which the polynomial l 3y³ - 3/2y²+ ky + 5. would actually be divisible by (y-1/2) = 0