Step 1

The quotient is x+5 and the remainder is -2

By using the formula,

Polynomial = Divisor \(\displaystyle\times\) Quotient + Remainder

Take divisor = x + 1

Thus, the polynomial is

Polynomial \(\displaystyle={\left({x}+{1}\right)}\times{\left({x}+{5}\right)}+{\left(-{2}\right)}\)

Polynomial \(\displaystyle={x}^{{2}}+{6}{x}+{5}-{2}\)

Polynomial \(\displaystyle={x}^{{2}}+{6}{x}+{3}\)

Step 2

By using the polynomial division,

\(\displaystyle\frac{{{x}^{{2}}+{6}{x}+{3}}}{{{x}+{1}}}\)

The first term is x2 so the first term in the quotient is x

\(\displaystyle{x}{\left({x}+{1}\right)}={x}^{{2}}+{x}\)

So, \(\displaystyle{x}^{{2}}+{6}{x}+{3}-{\left({x}^{{2}}+{x}\right)}={5}{x}+{3}\)

Now if 5x + 3 divide by x +1 then the quotient is x

5(x+1) = 5x + 5

So, 5x + 3 - (5x + 5) = -2 (This is the remainder)

The quotient is x+5 and remainder is -2

Hence, the solution

The quotient is x+5 and the remainder is -2

By using the formula,

Polynomial = Divisor \(\displaystyle\times\) Quotient + Remainder

Take divisor = x + 1

Thus, the polynomial is

Polynomial \(\displaystyle={\left({x}+{1}\right)}\times{\left({x}+{5}\right)}+{\left(-{2}\right)}\)

Polynomial \(\displaystyle={x}^{{2}}+{6}{x}+{5}-{2}\)

Polynomial \(\displaystyle={x}^{{2}}+{6}{x}+{3}\)

Step 2

By using the polynomial division,

\(\displaystyle\frac{{{x}^{{2}}+{6}{x}+{3}}}{{{x}+{1}}}\)

The first term is x2 so the first term in the quotient is x

\(\displaystyle{x}{\left({x}+{1}\right)}={x}^{{2}}+{x}\)

So, \(\displaystyle{x}^{{2}}+{6}{x}+{3}-{\left({x}^{{2}}+{x}\right)}={5}{x}+{3}\)

Now if 5x + 3 divide by x +1 then the quotient is x

5(x+1) = 5x + 5

So, 5x + 3 - (5x + 5) = -2 (This is the remainder)

The quotient is x+5 and remainder is -2

Hence, the solution