This note covers polynomial division, the remainder and factor theorems, and the relationships between the roots of a polynomial equation and its coefficients. Together they let you factorise cubics and beyond, find unknown coefficients, and work with roots you cannot easily calculate.
Key ideas
A polynomial of degree \(n\) has the form \(P(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0\) with \(a_n\ne 0\). Its roots are the values of \(x\) for which \(P(x)=0\). A polynomial of degree \(n\) has at most \(n\) real roots, and exactly \(n\) roots if complex roots are counted, with repeated roots counted each time.
Division. You can divide one polynomial by another, in the same way as long division of numbers. The result is written \(P(x)=D(x)\,Q(x)+R(x)\), where \(Q\) is the quotient and \(R\) is the remainder. The remainder has a lower degree than the divisor \(D\). When you divide by a linear factor, the remainder is a constant. You can divide using long division, or by comparing coefficients in an expression such as \(P(x)=(x-2)(x^2+px+q)\).
Remainder theorem. When \(P(x)\) is divided by \((x-a)\), the remainder is \(P(a)\). When it is divided by \((ax-b)\), the remainder is \(P\!\left(\dfrac ba\right)\). This is quick because you substitute and never divide.
Factor theorem. \((x-a)\) is a factor of \(P(x)\) if and only if \(P(a)=0\). In other words, \(a\) is a root exactly when \((x-a)\) divides \(P\) with no remainder. To factorise a cubic, test small whole-number values (and simple fractions) until one gives zero. Then divide to leave a quadratic, which you factorise or solve.
Sum and product of roots. For a polynomial equation \(a_nx^n+a_{n-1}x^{n-1}+\dots+a_0=0\) with roots \(\alpha_1,\alpha_2,\dots,\alpha_n\):
- the sum of the roots is \(-\dfrac{a_{n-1}}{a_n}\);
- the product of the roots is \((-1)^n\dfrac{a_0}{a_n}\).
For a quadratic \(ax^2+bx+c=0\), these give \(\alpha+\beta=-\dfrac ba\) and \(\alpha\beta=\dfrac ca\). For a cubic \(ax^3+bx^2+cx+d=0\), they give \(\alpha+\beta+\gamma=-\dfrac ba\), \(\alpha\beta+\beta\gamma+\gamma\alpha=\dfrac ca\) and \(\alpha\beta\gamma=-\dfrac da\). You can use these without knowing the roots, even if some are complex. Often you combine them with an identity such as \(\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)\).
Repeated and complex roots. If \((x-a)^2\) is a factor, then \(a\) is a repeated root and the graph touches the \(x\)-axis at \(x=a\) without crossing it. If \((x-a)^3\) is a factor, the graph flattens as it crosses. When the coefficients are real, any non-real roots come in conjugate pairs, so a cubic with real coefficients always has at least one real root. For example, \(x^3-2x^2+3x-6=(x-2)(x^2+3)\) has the real root \(x=2\) and the complex pair \(x=\pm i\sqrt3\).
Building an equation from its roots. If a quadratic has roots \(\alpha\) and \(\beta\), then it can be written \(x^2-(\alpha+\beta)x+\alpha\beta=0\). For example, roots 3 and \(-\tfrac12\) give a sum of \(\tfrac52\) and a product of \(-\tfrac32\), so \(x^2-\tfrac52x-\tfrac32=0\), or \(2x^2-5x-3=0\).
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(P(x)=D(x)Q(x)+R(x)\) | Result of polynomial division. \(R\) has lower degree than \(D\). |
| Remainder theorem: \(P(x)\div(x-a)\) leaves \(P(a)\) | Find a remainder or an unknown coefficient without dividing. |
| Factor theorem: \(P(a)=0\iff(x-a)\) is a factor | Test possible roots and factorise. |
| \(\displaystyle\sum\alpha_i=-\dfrac{a_{n-1}}{a_n}\) | Sum of all roots. |
| \(\displaystyle\prod\alpha_i=(-1)^n\dfrac{a_0}{a_n}\) | Product of all roots. The sign depends on the degree. |
| Cubic: \(\sum\alpha\beta=\dfrac ca\) | Sum of the roots taken two at a time. |
| \(\alpha^2+\beta^2+\gamma^2=(\sum\alpha)^2-2\sum\alpha\beta\) | Sum of squares of the roots of a cubic. |
Worked examples
Example 1 Let \(P(x)=2x^3-3x^2-11x+6\). Show that \((x+2)\) is a factor, then factorise \(P(x)\) completely and solve \(P(x)=0\).
Since \(x+2=x-(-2)\), evaluate \(P(-2)\): \(2(-8)-3(4)-11(-2)+6=-16-12+22+6=0\).
The remainder is 0, so by the factor theorem \((x+2)\) is a factor.
Write \(P(x)=(x+2)(2x^2+bx+3)\). The constant term must give \(2\times3=6\), and the \(x^3\) term gives 2, so this is the correct shape.
Compare the \(x^2\) terms: \(bx^2+4x^2=-3x^2\), so \(b+4=-3\) and \(b=-7\).
Check the \(x\) terms: \(3x+2bx=3x-14x=-11x\). This matches.
So \(P(x)=(x+2)(2x^2-7x+3)\). Factorise the quadratic: \(2x^2-7x+3=(2x-1)(x-3)\).
Therefore \(P(x)=(x+2)(2x-1)(x-3)\), and the roots are \(x=-2\), \(x=\tfrac12\) and \(x=3\).
Answer: \(P(x)=(x+2)(2x-1)(x-3)\); \(x=-2,\ \tfrac12,\ 3\)
Example 2 The polynomial \(P(x)=x^3+ax^2+bx-6\) has \((x-2)\) as a factor. When \(P(x)\) is divided by \((x+1)\), the remainder is \(-12\). Find \(a\) and \(b\).
The factor theorem gives \(P(2)=0\): \(8+4a+2b-6=0\), so \(4a+2b=-2\), which simplifies to \(2a+b=-1\). Call this equation (1).
The remainder theorem gives \(P(-1)=-12\): \(-1+a-b-6=-12\), so \(a-b=-5\). Call this equation (2).
Add (1) and (2): \(3a=-6\), so \(a=-2\).
Substitute into (2): \(-2-b=-5\), so \(b=3\).
Check (1): \(2(-2)+3=-1\). Correct.
Answer: \(a=-2\), \(b=3\)
Example 3 The roots of \(2x^3-5x^2+4x-7=0\) are \(\alpha\), \(\beta\) and \(\gamma\). Find \(\alpha+\beta+\gamma\), \(\alpha\beta+\beta\gamma+\gamma\alpha\), \(\alpha\beta\gamma\), and \(\alpha^2+\beta^2+\gamma^2\).
Here \(a=2\), \(b=-5\), \(c=4\) and \(d=-7\).
Sum of the roots: \(\alpha+\beta+\gamma=-\dfrac ba=-\dfrac{-5}{2}=\dfrac52\).
Sum in pairs: \(\alpha\beta+\beta\gamma+\gamma\alpha=\dfrac ca=\dfrac42=2\).
Product: \(\alpha\beta\gamma=-\dfrac da=-\dfrac{-7}{2}=\dfrac72\).
Use the identity: \(\alpha^2+\beta^2+\gamma^2=\left(\dfrac52\right)^2-2(2)=\dfrac{25}{4}-4=\dfrac94\).
No roots had to be found. (The equation has one real root and two complex roots, but the formulae still hold.)
Answer: \(\dfrac52\), \(2\), \(\dfrac72\), and \(\dfrac94\)
Common errors
- Using \(P(a)\) instead of \(P(-a)\) for a factor such as \((x+a)\). The factor \((x+2)\) corresponds to \(x=-2\).
- Forgetting the sign in the formulae for roots. The sum of the roots is \(-\dfrac{a_{n-1}}{a_n}\), so a positive \(x^2\) coefficient in a monic cubic gives a negative sum.
- Using the wrong sign for the product of the roots. For a quadratic and a quartic the product is \(+\dfrac{a_0}{a_n}\), but for a cubic it is \(-\dfrac{a_0}{a_n}\).
- For a divisor such as \((2x-1)\), writing the remainder as \(P(1)\). It is \(P\!\left(\tfrac12\right)\).
- Missing a zero coefficient when dividing. Write \(x^3-4\) as \(x^3+0x^2+0x-4\) first.
Exam tip: After finding a factor, expand your final factorised form quickly, or check one value of \(x\), to confirm it equals the original polynomial.
Check yourself
- Find the remainder when \(x^3-4x+5\) is divided by \((x-3)\).
Answer
By the remainder theorem, evaluate at \(x=3\): \(27-12+5=20\). - Show that \((x+1)\) is a factor of \(x^3+2x^2-5x-6\), and factorise fully.
Answer
At \(x=-1\): \(-1+2+5-6=0\), so \((x+1)\) is a factor. Dividing gives \(x^3+2x^2-5x-6=(x+1)(x^2+x-6)=(x+1)(x+3)(x-2)\). - Find the sum and product of the roots of \(2x^4-6x^3+x-8=0\).
Answer
Here \(a_4=2\), \(a_3=-6\), \(a_0=-8\). Sum \(=-\dfrac{-6}{2}=3\). Product \(=(-1)^4\dfrac{-8}{2}=-4\). - When \(x^3+kx^2-2x+5\) is divided by \((x-2)\) the remainder is 17. Find \(k\).
Answer
The remainder is \(P(2)=8+4k-4+5=4k+9\). Setting \(4k+9=17\) gives \(k=2\).