Question

ɪғ х³ + 8х² + ĸх+ 18 ɪs ᴄᴏᴍᴘʟᴇᴛʏ ᴅɪᴠɪsɪʙʟᴇ вy х² + 6х +9 ᴛʜᴇɴ ғɪɴᴅ ᴋ ​

Answer 1

Given:-

• $x^{3} + 8x^{2} + kx + 18$   is completely divisible by  $x^{2} + 6x + 9$ .

To Find:-

• Value of k .

Solution:-

Here, It is given that $x^{3} + 8x^{2} + kx + 18$   is completely divisible by  $x^{2} + 6x + 9$ .

So, Let f(x) = $x^{3} + 8x^{2} + kx + 18$

and, g(x) = $x^{2} + 6x + 9$

Now, From remainder theorem we have,

g(x) = $x^{2} + 6x + 9$ = 0

    ⇒ $x^{2} + 3x + 3x + 9 = 0$

    ⇒ $x (x + 3) + 3(x + 3)$ = 0

    ⇒ $(x + 3) (x + 3)$ = 0

    ⇒ $x = -3$

Now, putting the value of $x$ in f(x) we get,

f(-3) = $(-3)^{3} + 8(-3)^{2} + k(-3) + 18$ = 0

     ⇒ $-27 + 72 - 3k + 18$ = 0

     ⇒ $3k = 63$

     ⇒ $k = \dfrac{63}{3}$

     ⇒ k = 21

Hence, the value of k in $x^{3} + 8x^{2} + kx + 18$ is 21.

Some important terms:-

• Remainder Theorem :- According to this theorem, if we divide a polynomial P(x) by a factor ( x - a ); that isn’t essentially an element of the polynomial; you will find a smaller polynomial along with a remainder. This remainder that has been obtained is actually a value of P(x) at x = a, specifically P(a). So basically, x - a is the divisor of P(x) if and only if P(a) = 0.