Question

If the polynomial $g\left(x\right)=x^2+5x+k$ is a factor of $p\left(x\right)=4x^4+20x^3-26x^2-10x+12$ then find the value of 𝒌. Also find all the zeroes of the polynomial.

Answer 1

Step 1: Divide the highest degree term of the dividend by the highest degree term of the divisor.

here $4x^4$ is the highest term in dividend and $x^2$ is the highest term in divisor, so we get the first term of the quotient as $2x^2$

Step 2: Multiply the quotient with the divisor.

multiply the quotient obtained in the previous step with the divisor, hence the product is $4x^4+20x^3+4k{x}^2$

Step 3: Subtract the product of the divisor and the quotient from the dividend.

• we get $-(26+4k)a{x}^2-10x+12$ after subtraction
• Repeat the steps till the remainder is zero or deg r(x) < deg g(x).
• So, the quotient is $4x^2-(26+4k)$.
• And the remainder is $5(24+4k)x+k(26+4k)+12$

• Now, let’s equate the remainder to zero.
• $5x(24+4k)+k(26+4k)+12=0$
• 24 + 4k = 0

4k = -24
k = -6

• ​​​​​​$26k+4k^2+12=0$

$2k^2+13k+6=0$
$2k^2+12k+k+6=0$
$2k(k+6)+1(k+6)=0$
$(2k+1)(k+6)=0$
$k=-1/2,-6$

• As k = -6 is common in both so k = -6 is the value of k