Question

find the value of k if X + K is a factor of polynomial x cube + kx square - 2 X + K + 5​

Answer 1

Solution :-

Let f(x) = x³ + kx² - 2x + k + 5

x + k is a factor of x³ + kx² - 2x + k + 5

Finding the zero of x + k

⇒ x + k = 0

⇒ x = - k

By factor theorem

⇒ f(-k) = 0

⇒ (-k)³ + k(-k) ² - 2(-k) + k + 5 = 0

⇒ - k³ + k(k²) + 2k + k + 5 = 0

⇒ - k³ + k³ + 3k + 5 = 0

⇒ 3k + 5 = 0

⇒ 3k = - 5

⇒ k = - 5/3

Therefore the value of k is - 5/3.

Answer 2

ᴛʜᴇ ᴠᴀʟᴜᴇ ᴏғ 'ᴋ' ɪs $\bf\frac{-5}{3}$

$\textbf{\underline{\underline{Step \: by \: step \: explanation}}}$

ᴡᴇ ᴀʀᴇ ɢɪᴠᴇɴ ᴀ ᴘᴏʟʏɴᴏᴍɪᴀʟ ᴀɴᴅ ɪᴛs ғᴀᴄᴛᴏʀ ᴀɴᴅ ᴡᴇ ʜᴀᴠᴇ ᴛᴏ ғɪɴᴅ ᴛʜᴇ ᴠᴀʟᴜᴇ ᴏғ 'ᴋ'

ʟᴇᴛ's ᴄʟᴇᴀʀ ᴛʜᴇ ʙᴀsɪᴄ ᴄᴏɴᴄᴇᴘᴛ;

ᴡᴇ ᴀʀᴇ ɢᴏɪɴɢ ᴛᴏ sᴏʟᴠᴇ ᴏᴜʀ ǫᴜᴇsᴛɪᴏɴ ᴡɪᴛʜ ᴛʜᴇ ʜᴇʟᴘ ᴏғ ғᴀᴄᴛᴏʀ ᴛʜᴇᴏʀᴇᴍ

ʟᴇᴛ's,ʙᴇɢɪɴ ;

•ɢɪᴠᴇɴ ᴘᴏʟʏɴᴏᴍɪᴀʟ :

$\mapsto$$\normalsize\tt\ P(x) \: = \: x^3 \: + \: kx^2 \: - \: 2x \: + \: k \: + \: 5$

•ɢɪᴠᴇɴ ғᴀᴄᴛᴏʀ :

$\mapsto$$\normalsize\tt\ Factor \: = \: x \: + \: k$

ɴᴏᴛᴇ;

ᴀᴄᴄᴏʀᴅɪɴɢ ᴛᴏ ғᴀᴄᴛᴏʀ ᴛʜᴇᴏʀᴇᴍ,ɪғ x+ᴋ ɪs ᴀ ғᴀᴄᴛᴏʀ ᴏғ ɢɪᴠᴇɴ ᴘ(x),ᴛʜᴇɴ ᴡʜᴇɴ ɪᴛ ᴡɪʟʟ ʙᴇ ᴅɪᴠɪᴅᴇᴅ ʙʏ ᴘ(x) ɪᴛ ʟᴇᴀᴠᴇs 0 ᴀs ʀᴇᴍᴀɪɴᴅᴇʀ,

(ᴋᴇᴇᴘɪɴɢ ᴛʜɪs ɪɴ ᴍɪɴᴅ,ʟᴇᴛs sᴏʟᴠᴇ )

$\normalsize\mapsto\tt\ x\: + \: k \: = \: 0 \\ \\ \normalsize\mapsto\tt\ x \: = \: 0 \: - \: k$

$\mapsto$ ${\boxed{\sf{ x \: = \: -k}}}$

ɴᴏᴡ,sᴏʟᴠᴇ ʙʏ ʙʟᴏᴄᴋɪɴɢ ᴛʜᴇ ᴠᴀʟᴜᴇ ᴏғ 'ᴋ' ɪɴ ɢɪᴠᴇɴ ᴘᴏʟʏɴᴏᴍɪᴀʟ;

➠$\normalsize\tt\ P(-k) \: = \: (-k)^3 \: + \:k(-k)^2 \: - \: 2(-k) \: + \: k \: + \: 5$

➠$\normalsize\tt\ 0 \: = \: - \: k^3\cancel \: + \: k^3 \: + \: 2k \: + k \: + \: 5$

➠$\normalsize\tt\ 3k \: + \: 5 \: = \: 0$

➠$\normalsize\tt\ 3k \: = \: 0 \: - \: 5$

➠$\normalsize\tt\ 3k \: = \: -5$

➠$\normalsize\tt\ k \: = \: \frac{-5}{3}$

➠${\boxed{\tt \pink{Value \: of \: k \: = \: \frac{-5}{3} }}}$