Question

Find the value of 'k' if the polynomial p(x)=3x²-7x+2 leaves a remainder -1 when divided by (x+k).
pls solve this​

Answer 1

Answer:

( - 7 ± √13 ) / 6

Step-by-step explanation:

Given :

p( x ) = 3x² - 7x + 2 leaveas a reamainder - 1 when divided by ( x + k )

By reamainder theorem

p( - k ) will be the remainder when p( x ) is divided by ( x + k )

Also according to question - 1 is the remainder

So, let's equate both

⇒ p( - k ) = - 1

⇒ 3( - k )² - 7( - k ) + 2 = - 1

⇒ 3k² + 7k + 2 + 1 = 0

⇒ 3k² + 7k + 3 = 0

Using Quadratic formula

$\boxed{ \sf x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} }$

• a = 3
• b = 7
• c = 3

$\Rightarrow k = \dfrac{ - 7 \pm \sqrt{ {7}^{2} - 4(3)(3) } }{2(3)}$

$\Rightarrow k= \dfrac{ - 7 \pm \sqrt{ 49 - 36 } }{6}$

$\Rightarrow k= \dfrac{ - 7 \pm \sqrt{ 13 } }{6}$

Therefore the value of k is ( - 7 ± √13 ) / 6.

Answer 2

Answer:

Given : p( x) = 3x² – 7x + 2 leaves a remainder – 1 when divided by (x + k)

When we'll Subtract – 1 from the p(x) then it will be completely Divisible by (x + k).

☯ New Dividend will be :

$\dashrightarrow\sf\:\:p(x)=3x^2-7x + 2-(-1)\\\\\\\dashrightarrow\sf\:\:p(x)=3x^2-7x + 2 + 1\\\\\\\dashrightarrow\sf\:\:p(x)=3x^2-7x + 3$

Now p( x) = 3x² – 7x + 3 is completely Divisible by (x + k), & so – k is one of the Factor of p( x)

$\rule{110}{0.9}$

$\underline{\bigstar\:\sf{According\:to\:the\:Question :}}$

$:\implies\sf p(x)=3x^2-7x+3\\\\\\:\implies\sf p(-k)=3(-k)^2-7(-k)+3\\\\\\:\implies\sf 3k^2+7k+3=0\\\\{\scriptsize\qquad\bf{\dag}\:\:\texttt{Using Discriminat Formula.}}\\\\:\implies\sf k=\dfrac{-\:b\pm\sqrt{b^2-4ac}}{2a}\\\\\\:\implies\sf k=\dfrac{-\:7\pm\sqrt{(7)^2-(4 \times 3 \times 3)}}{2 \times 3}\\\\\\:\implies\sf k = \dfrac{-\:7\pm\sqrt{49-36}}{6}\\\\\\:\implies\underline{\boxed{\sf k =\dfrac{-\:7\pm\sqrt{13}}{6}}}$