Solved Exercises on Polynomial Long Division

\[ Q(x)=x+3,\quad R(x)=0 \]

Key idea

The dividend factors as \((x+2)(x+3)\): the division will be exact. The algorithm confirms this in just two steps.

Step 1

I divide the leading term: \(x^2\div x=x\). I multiply: \(x(x+2)=x^2+2x\). I change signs and add: \(x^2\) cancels. Remaining polynomial: \(3x+6\).

Step 2

I divide: \(3x\div x=3\). I multiply: \(3(x+2)=3x+6\). I change signs: \(3x\) and \(6\) cancel. Remainder: \(0\).

Full long-division layout

Result

\[ \boxed{Q(x)=x+3,\quad R(x)=0} \]

Check: \( (x+2)(x+3)=x^2+5x+6\;\checkmark \)

\[ Q(x)=x+3,\quad R(x)=0 \]

Key idea

We recognise the form \(a^2-b^2=(a-b)(a+b)\) with \(a=x\) and \(b=3\). The term \(0x\) must be inserted as a placeholder.

Step 1

I divide: \(x^2\div x=x\). I multiply: \(x(x-3)=x^2-3x\). I change signs: \(x^2\) cancels. Remaining: \(3x-9\).

Step 2

I divide: \(3x\div x=3\). I multiply: \(3(x-3)=3x-9\). I change signs: \(3x\) and \(-9\) cancel. Remainder: \(0\).

Full long-division layout

Result

\[ \boxed{Q(x)=x+3,\quad R(x)=0} \]

Check: \( (x-3)(x+3)=x^2-9\;\checkmark \)

\[ Q(x)=x+3,\quad R(x)=0 \]

Key idea

Since \(f(1)=1+2-3=0\), the remainder theorem guarantees that \((x-1)\) divides the dividend exactly.

Step 1

I divide: \(x^2\div x=x\). I multiply: \(x(x-1)=x^2-x\). I change signs: \(x^2\) cancels. Remaining: \(3x-3\).

Step 2

I divide: \(3x\div x=3\). I multiply: \(3(x-1)=3x-3\). I change signs: everything cancels. Remainder: \(0\).

Full long-division layout

Result

\[ \boxed{Q(x)=x+3,\quad R(x)=0} \]

Check: \( (x-1)(x+3)=x^2+2x-3\;\checkmark \)

\[ Q(x)=2x-3,\quad R(x)=0 \]

Key idea

The leading coefficient of the dividend is \(2\): the first term of the quotient will be \(2x\). The division is exact because \(f(-1)=0\).

Step 1

I divide: \(2x^2\div x=2x\). I multiply: \(2x(x+1)=2x^2+2x\). I change signs: \(2x^2\) cancels. Remaining: \(-3x-3\).

Step 2

I divide: \(-3x\div x=-3\). I multiply: \(-3(x+1)=-3x-3\). I change signs: everything cancels. Remainder: \(0\).

Full long-division layout

Result

\[ \boxed{Q(x)=2x-3,\quad R(x)=0} \]

Check: \( (x+1)(2x-3)=2x^2-x-3\;\checkmark \)

\[ Q(x)=x+1,\quad R(x)=2 \]

Key idea

By the remainder theorem, \(R=f(1)=1+1=2\neq0\): the division is not exact. The term \(0x\) must be inserted as a placeholder.

Step 1

I divide: \(x^2\div x=x\). I multiply: \(x(x-1)=x^2-x\). I change signs: \(x^2\) cancels. Remaining: \(x+1\).

Step 2

I divide: \(x\div x=1\). I multiply: \(1\cdot(x-1)=x-1\). I change signs: \(x\) cancels; \(1+1=2\). Since \(\deg 0<1\), we stop.

Full long-division layout

Result

\[ \boxed{Q(x)=x+1,\quad R(x)=2} \]

Check: \( (x-1)(x+1)+2=x^2+1\;\checkmark \)

\[ Q(x)=x^2+2x+4,\quad R(x)=0 \]

Key idea

Formula: \(a^3-b^3=(a-b)(a^2+ab+b^2)\) with \(a=x,\;b=2\). The terms \(0x^2\) and \(0x\) must be inserted as placeholders.

Step 1

I divide: \(x^3\div x=x^2\). I multiply: \(x^2(x-2)=x^3-2x^2\). I change signs: \(x^3\) cancels. Remaining: \(2x^2-8\).

Step 2

I divide: \(2x^2\div x=2x\). I multiply: \(2x(x-2)=2x^2-4x\). I change signs: \(2x^2\) cancels. Remaining: \(4x-8\).

Step 3

I divide: \(4x\div x=4\). I multiply: \(4(x-2)=4x-8\). I change signs: everything cancels. Remainder: \(0\).

Full long-division layout

Result

\[ \boxed{Q(x)=x^2+2x+4,\quad R(x)=0} \]

Check: \( (x-2)(x^2+2x+4)=x^3-8\;\checkmark \)

\[ Q(x)=x^2-x+1,\quad R(x)=0 \]

Key idea

Formula: \(a^3+b^3=(a+b)(a^2-ab+b^2)\) with \(a=x,\;b=1\). Quick check: \(f(-1)=-1+1=0\).

Step 1

I divide: \(x^3\div x=x^2\). I multiply: \(x^2(x+1)=x^3+x^2\). I change signs: \(x^3\) cancels. Remaining: \(-x^2+1\).

Step 2

I divide: \(-x^2\div x=-x\). I multiply: \(-x(x+1)=-x^2-x\). I change signs: \(-x^2\) cancels. Remaining: \(x+1\).

Step 3

I divide: \(x\div x=1\). I multiply: \(1\cdot(x+1)=x+1\). I change signs: everything cancels. Remainder: \(0\).

Full long-division layout

Result

\[ \boxed{Q(x)=x^2-x+1,\quad R(x)=0} \]

Check: \( (x+1)(x^2-x+1)=x^3+1\;\checkmark \)

\[ Q(x)=x^2+x+2,\quad R(x)=3 \]

Key idea

The \(x^2\) term is missing: insert it as \(0x^2\). By the remainder theorem, \(f(1)=1+1+1=3\neq0\), so the remainder is \(3\).

Step 1

I divide: \(x^3\div x=x^2\). I multiply: \(x^2(x-1)=x^3-x^2\). I change signs: \(x^3\) cancels. Remaining: \(x^2+x+1\).

Step 2

I divide: \(x^2\div x=x\). I multiply: \(x(x-1)=x^2-x\). I change signs: \(x^2\) cancels. Remaining: \(2x+1\).

Step 3

I divide: \(2x\div x=2\). I multiply: \(2(x-1)=2x-2\). I change signs: \(2x\) cancels; \(1+2=3\). Since \(\deg 0<1\), we stop.

Full long-division layout

Result

\[ \boxed{Q(x)=x^2+x+2,\quad R(x)=3} \]

Check: \( (x-1)(x^2+x+2)+3=x^3+x+1\;\checkmark \)

\[ Q(x)=x^2-2x+1,\quad R(x)=0 \]

Key idea

The dividend is \((x-1)^3\). Dividing by \((x-1)\) gives \((x-1)^2=x^2-2x+1\). Check: \(f(1)=0\).

Step 1

I divide: \(x^3\div x=x^2\). I multiply: \(x^2(x-1)=x^3-x^2\). I change signs: \(x^3\) cancels. Remaining: \(-2x^2+3x-1\).

Step 2

I divide: \(-2x^2\div x=-2x\). I multiply: \(-2x(x-1)=-2x^2+2x\). I change signs: \(-2x^2\) cancels. Remaining: \(x-1\).

Step 3

I divide: \(x\div x=1\). I multiply: \(1\cdot(x-1)=x-1\). I change signs: everything cancels. Remainder: \(0\).

Full long-division layout

Result

\[ \boxed{Q(x)=x^2-2x+1,\quad R(x)=0} \]

Check: \( (x-1)^3\div(x-1)=(x-1)^2=x^2-2x+1\;\checkmark \)

\[ Q(x)=x-1,\quad R(x)=-1 \]

Key idea

By the remainder theorem, \(R=f(2)=4-6+1=-1\neq0\). The quotient has degree \(1\).

Step 1

I divide: \(x^2\div x=x\). I multiply: \(x(x-2)=x^2-2x\). I change signs: \(x^2\) cancels. Remaining: \(-x+1\).

Step 2

I divide: \(-x\div x=-1\). I multiply: \(-1\cdot(x-2)=-x+2\). I change signs: \(-x\) cancels; \(1-2=-1\). Since \(\deg 0<1\), we stop.

Full long-division layout

Result

\[ \boxed{Q(x)=x-1,\quad R(x)=-1} \]

Check: \( (x-2)(x-1)-1=x^2-3x+1\;\checkmark \)

\[ Q(x)=x^2+2x+1,\quad R(x)=0 \]

Key idea

\(f(1)=1+1-1-1=0\): the division is exact. The quotient \(x^2+2x+1=(x+1)^2\) is a perfect square.

Step 1

I divide: \(x^3\div x=x^2\). I multiply: \(x^2(x-1)=x^3-x^2\). I change signs: \(x^3\) cancels. Remaining: \(2x^2-x-1\).

Step 2

I divide: \(2x^2\div x=2x\). I multiply: \(2x(x-1)=2x^2-2x\). I change signs: \(2x^2\) cancels. Remaining: \(x-1\).

Step 3

I divide: \(x\div x=1\). I multiply: \(1\cdot(x-1)=x-1\). I change signs: everything cancels. Remainder: \(0\).

Full long-division layout

Result

\[ \boxed{Q(x)=x^2+2x+1,\quad R(x)=0} \]

Check: \( (x-1)(x+1)^2=x^3+x^2-x-1\;\checkmark \)

\[ Q(x)=2x^2-4x+5,\quad R(x)=-9 \]

Key idea

The \(x^2\) term is missing: insert it as \(0x^2\). The remainder theorem gives \(f(-2)=-16+6+1=-9\): this confirms the remainder.

Step 1

I divide: \(2x^3\div x=2x^2\). I multiply: \(2x^2(x+2)=2x^3+4x^2\). I change signs: \(2x^3\) cancels. Remaining: \(-4x^2-3x+1\).

Step 2

I divide: \(-4x^2\div x=-4x\). I multiply: \(-4x(x+2)=-4x^2-8x\). I change signs: \(-4x^2\) cancels. Remaining: \(5x+1\).

Step 3

I divide: \(5x\div x=5\). I multiply: \(5(x+2)=5x+10\). I change signs: \(5x\) cancels; \(1-10=-9\). Since \(\deg 0<1\), we stop.

Full long-division layout

Result

\[ \boxed{Q(x)=2x^2-4x+5,\quad R(x)=-9} \]

Check: \( (x+2)(2x^2-4x+5)-9=2x^3-3x+1\;\checkmark \)

\[ Q(x)=x+2,\quad R(x)=0 \]

Key idea

The divisor \(x^2-1=(x-1)(x+1)\) has degree 2: the quotient will have degree \(3-2=1\) and the remainder at most degree \(1\).

Step 1

I divide: \(x^3\div x^2=x\). I multiply: \(x(x^2-1)=x^3-x\). I change signs: \(x^3\) and \(-x\) cancel. Remaining: \(2x^2-2\).

Step 2

I divide: \(2x^2\div x^2=2\). I multiply: \(2(x^2-1)=2x^2-2\). I change signs: everything cancels. Remainder: \(0\).

Full long-division layout

Result

\[ \boxed{Q(x)=x+2,\quad R(x)=0} \]

Check: \( (x^2-1)(x+2)=x^3+2x^2-x-2\;\checkmark \)

\[ Q(x)=2x+1,\quad R(x)=-2x+8 \]

Key idea

The divisor has degree 2 and the dividend degree 3: the quotient will have degree \(1\). The remainder has degree at most \(1\), i.e., it is of the form \(ax+b\).

Step 1

I divide: \(2x^3\div x^2=2x\). I multiply: \(2x(x^2-x-2)=2x^3-2x^2-4x\). I change signs: \(2x^3\) cancels. Remaining: \(x^2-3x+6\).

Step 2

I divide: \(x^2\div x^2=1\). I multiply: \(x^2-x-2\). I change signs: \(x^2\) cancels. Remaining: \(-2x+8\). Since \(\deg 1<2\), we stop.

Full long-division layout

Result

\[ \boxed{Q(x)=2x+1,\quad R(x)=-2x+8} \]

Check: \( (x^2-x-2)(2x+1)+(-2x+8)=2x^3-x^2-7x+6\;\checkmark \)

\[ Q(x)=x^2-x+2,\quad R(x)=-3 \]

Key idea

The \(x^3\) term is missing: insert it as \(0x^3\). The quotient will have degree \(4-2=2\). The remainder is a constant.

Step 1

I divide: \(x^4\div x^2=x^2\). I multiply: \(x^2(x^2+x+1)=x^4+x^3+x^2\). I change signs: \(x^4\) and \(x^3\) cancel. Remaining: \(-x^3+x^2+x-1\).

Step 2

I divide: \(-x^3\div x^2=-x\). I multiply: \(-x(x^2+x+1)=-x^3-x^2-x\). I change signs: \(-x^3\), \(x^2\) and \(x\) cancel. Remaining: \(2x^2+2x-1\).

Step 3

I divide: \(2x^2\div x^2=2\). I multiply: \(2(x^2+x+1)=2x^2+2x+2\). I change signs: \(2x^2\) and \(2x\) cancel; \(-1-2=-3\). Since \(\deg 0<2\), we stop.

Full long-division layout

Result

\[ \boxed{Q(x)=x^2-x+2,\quad R(x)=-3} \]

Check: \( (x^2+x+1)(x^2-x+2)-3=x^4+2x^2+x-1\;\checkmark \)

\[ Q(x)=x^2-4,\quad R(x)=0 \]

Key idea

The terms \(x^3\) and \(x\) are missing: insert them as \(0x^3\) and \(0x\). We recognise \(x^4-5x^2+4=(x^2-1)(x^2-4)\).

Step 1

I divide: \(x^4\div x^2=x^2\). I multiply: \(x^2(x^2-1)=x^4-x^2\). I change signs: \(x^4\) cancels. Remaining: \(-4x^2+4\).

Step 2

I divide: \(-4x^2\div x^2=-4\). I multiply: \(-4(x^2-1)=-4x^2+4\). I change signs: everything cancels. Remainder: \(0\).

Full long-division layout

Result

\[ \boxed{Q(x)=x^2-4,\quad R(x)=0} \]

Check: \( (x^2-1)(x^2-4)=x^4-5x^2+4\;\checkmark \)

\[ Q(x)=3x-5,\quad R(x)=9x-9 \]

Key idea

Dividend degree 3, divisor degree 2: quotient of degree \(1\), remainder of degree at most \(1\). The remainder is non-zero and must be fully computed.

Step 1

I divide: \(3x^3\div x^2=3x\). I multiply: \(3x(x^2+x-1)=3x^3+3x^2-3x\). I change signs: \(3x^3\) cancels. Remaining: \(-5x^2+4x-4\).

Step 2

I divide: \(-5x^2\div x^2=-5\). I multiply: \(-5(x^2+x-1)=-5x^2-5x+5\). I change signs: \(-5x^2\) cancels. Remaining: \(9x-9\). Since \(\deg 1<2\), we stop.

Full long-division layout

Result

\[ \boxed{Q(x)=3x-5,\quad R(x)=9x-9} \]

Check: \( (x^2+x-1)(3x-5)+(9x-9)=3x^3-2x^2+x-4\;\checkmark \)

\[ Q(x)=x^2-2x-3,\quad R(x)=0 \]

Key idea

The divisor \(x^2+x-2=(x-1)(x+2)\). Both \(f(1)\) and \(f(-2)\) are zero: the division is exact. The quotient is itself factorable.

Step 1

I divide: \(x^4\div x^2=x^2\). I multiply: \(x^2(x^2+x-2)=x^4+x^3-2x^2\). I change signs: \(x^4\) cancels. Remaining: \(-2x^3-5x^2+x+6\).

Step 2

I divide: \(-2x^3\div x^2=-2x\). I multiply: \(-2x(x^2+x-2)=-2x^3-2x^2+4x\). I change signs: \(-2x^3\) cancels. Remaining: \(-3x^2-3x+6\).

Step 3

I divide: \(-3x^2\div x^2=-3\). I multiply: \(-3(x^2+x-2)=-3x^2-3x+6\). I change signs: everything cancels. Remainder: \(0\).

Full long-division layout

Result

\[ \boxed{Q(x)=x^2-2x-3,\quad R(x)=0} \]

Check: \( (x^2+x-2)(x^2-2x-3)=x^4-x^3-7x^2+x+6\;\checkmark \)

\[ Q(x)=x^4+x^3+x^2+x+1,\quad R(x)=0 \]

Key idea

Geometric series identity: \(\displaystyle\frac{x^5-1}{x-1}=x^4+x^3+x^2+x+1\). All intermediate terms of the dividend are zero.

Step 1

I divide: \(x^5\div x=x^4\). I multiply: \(x^4(x-1)=x^5-x^4\). I change signs: \(x^5\) cancels. Remaining: \(x^4-1\).

Step 2

I divide: \(x^4\div x=x^3\). I multiply: \(x^3(x-1)=x^4-x^3\). I change signs: \(x^4\) cancels. Remaining: \(x^3-1\).

Step 3

I divide: \(x^3\div x=x^2\). I multiply: \(x^2(x-1)=x^3-x^2\). I change signs: \(x^3\) cancels. Remaining: \(x^2-1\).

Step 4

I divide: \(x^2\div x=x\). I multiply: \(x(x-1)=x^2-x\). I change signs: \(x^2\) cancels. Remaining: \(x-1\).

Step 5

I divide: \(x\div x=1\). I multiply: \(1\cdot(x-1)=x-1\). I change signs: everything cancels. Remainder: \(0\).

Full long-division layout

Result

\[ \boxed{Q(x)=x^4+x^3+x^2+x+1,\quad R(x)=0} \]

Check: \( (x-1)(x^4+x^3+x^2+x+1)=x^5-1\;\checkmark \)

\[ Q(x)=x^2-1,\quad R(x)=0 \]

Key idea

The dividend factors as \(x(x-2)(x^2-1)\) and the divisor as \(x(x-2)\): the division is exact in just two steps.

Step 1

I divide: \(x^4\div x^2=x^2\). I multiply: \(x^2(x^2-2x)=x^4-2x^3\). I change signs: \(x^4\) and \(-2x^3\) cancel. Remaining: \(-x^2+2x\).

Step 2

I divide: \(-x^2\div x^2=-1\). I multiply: \(-1\cdot(x^2-2x)=-x^2+2x\). I change signs: everything cancels. Remainder: \(0\).

Full long-division layout

Result

\[ \boxed{Q(x)=x^2-1,\quad R(x)=0} \]

Check: \( x(x-2)(x^2-1)=x(x-2)(x-1)(x+1)\;\checkmark \)