Question

Any Real Genius Here??
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1⃣If the remainder division of 
${x}^{3} + 2 {x}^{2} + kx + 3$
by 
$x - 3$
is 21,Find the quotient and value of k
Hence find the zeros of cubic polynomial 
${x}^{3} + 2 {x}^{2} + kx - 18$

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Answer 1

Step-by-step explanation:

Let p(x) = x³ + 2x² + kx + 3.

On dividing f(x) by x - 3, the remainder is 21.

Then, f(3) = 21

⇒ (3)³ + 2(3)² + k(3) + 3 = 21

⇒ 27 + 18 + 3k + 3 = 21

⇒ 3k + 48 = 21

⇒ 3k = -27

⇒ k = -9

∴ p(x) = x³ + 2x² - 9x + 3.

By division algorithm, we know that

p(x) = g(x) * q(x) + r(x)

⇒ x³ + 2x² - 9x + 3 = (x - 3) * q(x) + 21.

⇒ x³ + 2x² - 9x - 18 = (x - 3) * q(x)

⇒ (x³ + 2x² - 9x - 18) ÷ (x - 3) = q(x)

Long Division method:

x - 3) x³ + 2x² - 9x - 18 (x² + 5x + 6

        x³ - 3x²

        -------------------------

                5x² - 9x - 18

                5x² - 15x

            ----------------------

                            6x - 18

                            6x - 18

         

                ---------------------

                                    0

i.e x³ + 2x² + kx - 18 = (x - 3)(x² + 5x + 6)

For zeroes,

⇒ (x - 3)(x² + 5x + 6) = 0

⇒ (x - 3)(x² + 2x + 3x + 6) = 0

⇒ (x - 3)(x(x + 2) + 3(x + 2)) = 0

⇒ (x - 3)(x + 2)(x + 3) = 0

⇒ x = 3, -2, -3.

Therefore:

Quotient = x² + 5x + 6

Value of k = -9

Zeroes of cubic polynomial = 3, -2, -3.

Hope it helps!

Answer 2

Dividend=Divisor×Quotient+Remainder

Given that,

Dividend=x³+2x²+kx+3

Divisor=x-3

Remainder=21

Now we have,

x³+2x²+kx+3=(x-3)quotient+21

(x³+2x²+kx-18)/(x-3)=quotient

Now the remainder will be equal to 0.

From above picture we get,

Remainder=0

3(k+15)-18=0

k+15=6

k=-9

Now eq1 we have,

x³+2x²-9x+3=(x-3)(x²+5x+6)+21

x³+2x²-9x-18=(x-3)(x²+2x+3x+6)

x³+2x²-9x-18=(x-3){x(x+2)+3(x+2)}

x³+2x²-9x-18=(x-3)(x+3)(x+2)

Therefore zeroes are 3 , -3 and -2