Question

What must be added to f(x) = 6x^4 + 8x^3 + 18x^2 +
20x + 5 so that the resulting polynomial is divisible
by g(x) = 3x^2 + 2x + 1?
plz solution bata dena please

Answer 1

Answer:

Step-by-step explanation

3x^2+2x+1

3x^2-3x+1x+1

3x(x-1)-1(x-1)

(3x-1) (x-1)

X=1/3;X=1

F(x)=6x^4+8x^3+18x^2+20x+5

F(1)=6(1)^4+8(1)^3+18(1)^2+20(1)+5

6+8+18+20+5

=57.

57 must be added to the given polynomial 6x^4+8x^3+18x^2+20x+5

Answer 2

Given:

f(x) = 6x^4 + 8x^3 + 18x^2 +  20x + 5

g(x) = 3x^2 + 2x + 1

To find:

What must be added to f(x) = 6x^4 + 8x^3 + 18x^2 +  20x + 5 so that the resulting polynomial is divisible  by g(x) = 3x^2 + 2x + 1?

Solution:

We use the relation: dividend = divisor × quotient + remainder

f(x) = g(x) × q(x) + r(x)

$\dfrac{6x^4+8x^3+18x^2+20x+5}{3x^2+2x+1}$

$=2x^2+\dfrac{4x^3+16x^2+20x+5}{3x^2+2x+1}$

$=2x^2+\dfrac{4x}{3}+\dfrac{\frac{40x^2}{3}+\frac{56x}{3}+5}{3x^2+2x+1}$

$=2x^2+\dfrac{4x}{3}+\dfrac{40}{9}+\dfrac{\frac{88x}{9}+\frac{5}{9}}{3x^2+2x+1}$

Quotient = q(x) = $2x^2+\dfrac{4x}{3}+\dfrac{40}{9}$

Remainder = r(x) = $\dfrac{88x}{9}+\dfrac{5}{9}$

Therefore, $\dfrac{88x}{9}+\dfrac{5}{9}$ must be added to the polynomial f(x) = 6x^4 + 8x^3 + 18x^2 +  20x + 5 so that the resulting polynomial is divisible  by g(x) = 3x^2 + 2x + 1.