Question

If x3
– 7x2 + kx + 15 is exactly divisible by (x – 3), then find the value of k.​

Answer 1

Step-by-step explanation:

Let, f(x)=x^3-7x^2+kx+15

f(x) is divisible by (x-3)

then, x-3=0

or, x=3

So, f(3)=0

or, 3^3-7×3^2+k×3+15=0

or, 27-63+3k+15=0

or, 3k-21=0

or, 3k=21

or, k=21/3

or, k=7

Answer 2

SOLUTION

GIVEN

x³ - 7x² + kx + 15 is exactly divisible by ( x - 3 )

TO DETERMINE

The value of k

EVALUATION

Let f(x) = x³ - 7x² + kx + 15

g(x) = x - 3

For Zero of the polynomial g(x) we have

g(x) = 0

$\implies \sf{x - 3 = 0}$

$\implies \sf{x = 3}$

So by the Remainder Theorem the required Remainder when f(x) is divided by g(x)

$= \sf{f(3)}$

$= \sf{ {(3)}^{3} - 7 \times {(3)}^{2} + (k \times 3) + 15}$

$= \sf{ 27 - 63 + 3k + 15}$

$= \sf{ 3k - 21}$

Since x³ - 7x² + kx + 15 is exactly divisible by ( x - 3 )

∴ Remainder = 0

$\implies \sf{3k - 21 = 0}$

$\implies \sf{3k = 21}$

$\implies \sf{k = 7}$

FINAL ANSWER

The required value of k = 7

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