4.3 Division of polynomials
Dividing one polynomial by another works like long division of numbers, and produces the same two outputs: a quotient and a remainder. In symbols, \[P(x)=Q(x)\,D(x)+R(x),\] where \(D\) is the divisor, \(Q\) the quotient and \(R\) the remainder, whose degree is less than that of \(D\).
Solution. Set it out as long division, dealing with the highest power at each stage.
Step 1. \(2x^3\div 2x^2=x\). Multiply the divisor by \(x\) and subtract: \[x\left (2x^2-2\right )=2x^3-2x\] \[\left (2x^3+6x^2-x+3\right )-\left (2x^3-2x\right )=6x^2+x+3.\]
Step 2. \(6x^2\div 2x^2=3\). Multiply and subtract again: \[3\left (2x^2-2\right )=6x^2-6\] \[\left (6x^2+x+3\right )-\left (6x^2-6\right )=x+9.\]
The remainder \(x+9\) has degree \(1\), lower than the divisor’s degree \(2\), so we stop: \[\frac {2x^3+6x^2-x+3}{2x^2-2}=x+3+\frac {x+9}{2x^2-2}.\]
Check. \((x+3)(2x^2-2)+(x+9)=2x^3-2x+6x^2-6+x+9=2x^3+6x^2-x+3\), the original.
Note 4.19. Subtracting a bracket changes every sign inside it. Writing the subtraction out in full before simplifying, as above, is slower but far more reliable than doing it in the head.
4.3.1 The remainder and factor theorems
When the divisor is linear there is a shortcut that avoids the division entirely.
Note 4.20. Remainder theorem. When \(P(x)\) is divided by \(x-a\), the remainder is \(P(a)\).
Factor theorem. Consequently \(x-a\) is a factor of \(P(x)\) exactly when \(P(a)=0\).
The reason is short. Writing \(P(x)=Q(x)(x-a)+R\) with \(R\) constant, and putting \(x=a\), the first term vanishes and \(P(a)=R\).
Example 4.21. Divide \(P(x)=x^3-7x^2-4x+28\) by \(x-3\), and verify the remainder using the remainder theorem.
Solution. Synthetic division lists only the coefficients. Write them in order — \(1\), \(-7\), \(-4\), \(28\) — with \(a=3\) outside. Bring down the first, then repeatedly multiply by \(3\) and add.
| \(3\) | \(1\) | \(-7\) | \(-4\) | \(28\) |
| \(3\) | \(-12\) | \(-48\) | ||
| \(1\) | \(-4\) | \(-16\) | \(-20\) | |
So \[x^3-7x^2-4x+28=\left (x^2-4x-16\right )(x-3)-20.\]
Verification. By the remainder theorem the remainder should be \(P(3)\): \[P(3)=27-7(9)-4(3)+28=27-63-12+28=-20. \checkmark \] Since the remainder is not zero, \(x-3\) is not a factor.
Solution. Find one root by trial. Any integer root must divide the constant term \(28\), so the candidates are \(\pm 1,\pm 2,\pm 4,\pm 7,\pm 14,\pm 28\). Try the small ones: \[P(1)=1-7-4+28=18\neq 0,\qquad P(2)=8-28-8+28=0.\ \checkmark \] So by the factor theorem \(x-2\) is a factor.
Divide it out. Synthetic division by \(x-2\) gives \[P(x)=(x-2)\left (x^2-5x-14\right ).\]
Factorise the quadratic. Two numbers multiplying to \(-14\) and adding to \(-5\) are \(-7\) and \(2\): \[x^2-5x-14=(x-7)(x+2).\]
\[\therefore \quad P(x)=(x-2)(x-7)(x+2),\] and the zeros are \(x=2\), \(x=7\) and \(x=-2\).
Check. The three roots multiply to \(2\times 7\times (-2)=-28\), and for a cubic \(x^3+\cdots +a_0\) the product of the roots is \(-a_0=-28\). They agree.
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