Polynomial manipulation and algebraic divisionAQA A-Level Maths: Revision notes

A polynomial is a sum of terms of the form $ax^n$ with $n$ a whole number. The degree is the highest power of $x$ and the coefficient is the number multiplying a power of $x$, so $2x^3-x^2-13x-6$ has degree 3, leading coefficient 2 and constant term $-6$. To expand brackets, multiply every term in one bracket by every term in the other, then collect like terms (terms with the same power of $x$). For three brackets, expand two first and then multiply by the third. Example: $(x+2)(x-3)=x^2-x-6$, so $(x+2)(x-3)(2x+1)=(x^2-x-6)(2x+1)=2x^3+x^2-2x^2-x-12x-6=2x^3-x^2-13x-6$. To add or subtract polynomials, combine matching powers. For $p(x)-q(x)$ remember that subtracting a negative term adds it.

Factorising reverses expanding. Always look for a common factor first, e.g. $6x^3+9x^2=3x^2(2x+3)$. Difference of two squares: $a^2-b^2=(a-b)(a+b)$, e.g. $4x^2-25=(2x-5)(2x+5)$, and $x^4-81=(x^2-9)(x^2+9)=(x-3)(x+3)(x^2+9)$. Always check whether a factor can be factorised again; $x^2+9$ cannot. Quadratics $ax^2+bx+c$: find two numbers with product $ac$ and sum $b$, split the middle term, then factorise in pairs. For $6x^2+7x-3$, $ac=-18$ and the numbers are $9$ and $-2$: $6x^2+9x-2x-3=3x(2x+3)-1(2x+3)=(3x-1)(2x+3)$. Check by expanding.

When there are four terms, try grouping: factorise the first two terms and the last two terms and look for a common bracket. Example: $x^3+3x^2-4x-12=x^2(x+3)-4(x+3)=(x+3)(x^2-4)=(x+3)(x-2)(x+2)$. Grouping only works when the same bracket appears in both pairs. If it does not, rearrange the order of the terms or use algebraic division. Some expressions have a common bracket that is easy to miss. For $(2x-1)^2(x+3)-(x+3)^3$, take out $(x+3)$ first: $(x+3)\left[(2x-1)^2-(x+3)^2\right]$, then use the difference of two squares to get $(x+3)(x-4)(3x+2)$. This is quicker than expanding everything.

To divide a polynomial by $(x-a)$, use long division, working with the highest powers first. Write the polynomial in descending powers, including zero terms for missing powers such as $0x^2$. Example: $\frac{3x^3+2x^2-19x+6}{x-2}$.

• $3x^3\div x=3x^2$; subtract $3x^2(x-2)=3x^3-6x^2$, leaving $8x^2-19x$.
• $8x^2\div x=8x$; subtract $8x(x-2)=8x^2-16x$, leaving $-3x+6$.
• $-3x\div x=-3$; subtract $-3(x-2)=-3x+6$, leaving $0$. The quotient is $3x^2+8x-3$ and the remainder is $0$, so $3x^3+2x^2-19x+6=(x-2)(3x^2+8x-3)$.

If the remainder is not zero, write $f(x)=(x-a)\,q(x)+r$. For example, $2x^3-5x^2+4x-7$ divided by $(x-3)$ gives quotient $2x^2+x+7$ and remainder $14$, so $2x^3-5x^2+4x-7=(x-3)(2x^2+x+7)+14$. You can also divide by $(ax-b)$ in the same way: $\frac{6x^2+x-2}{2x-1}$ gives $3x+2$, because $6x^2\div2x=3x$, $3x(2x-1)=6x^2-3x$, leaving $4x-2$, then $4x\div2x=2$ with remainder $0$. Compare coefficients is an alternative: write $f(x)=(x-1)(2x^2+bx+c)$ and match powers of $x$. Always check by multiplying the quotient by the divisor and adding the remainder. After division, factorise the quotient if possible, e.g. $3x^2+8x-3=(3x-1)(x+3)$, to factorise the whole polynomial.