Algebraic Identities and Factorisation

Algebraic fraction simplification

An algebraic fraction is a quotient \frac{P(x)}{Q(x)} of two polynomials, valid only where Q(x)\neq 0. Simplifying it means factorising the numerator and the denominator completely and then cancelling the factors they share. Only multiplicative factors may be cancelled, because a term that is added or subtracted cannot be removed from a sum, so \frac{x^2+1}{x+1} cannot be reduced by deleting the x^2 and the 1. By contrast, \frac{x^2+2x}{x^2+x}=\frac{x(x+2)}{x(x+1)}=\frac{x+2}{x+1} after the common factor x is cancelled. The identities used most often are a^2-b^2=(a-b)(a+b), a^3-b^3=(a-b)(a^2+ab+b^2) and a^3+b^3=(a+b)(a^2-ab+b^2). A quadratic is factorised by finding two numbers whose product is the constant term and whose sum is the coefficient of x, as in x^2-5x+6=(x-2)(x-3). A perfect square such as x^2+4\sqrt{3}x+12=(x+2\sqrt{3})^2 must also be recognised, because it can cancel against a squared factor elsewhere. When fractions are added or subtracted, bring them to a common denominator built from the factored denominators, and then simplify the numerator. A missing term is found the same way, by subtracting the known fraction from the target: \frac{2}{x^2-1}-\frac{1}{x-1}=\frac{2-(x+1)}{x^2-1}=\frac{1-x}{x^2-1}=-\frac{1}{x+1}, so recognising x^2-1=(x-1)(x+1) is the decisive step. Symmetric sums with cyclic denominators need a systematic check. Writing \frac{1}{bc(a-b)(a-c)}+\frac{1}{ca(b-c)(b-a)}+\frac{1}{ab(c-a)(c-b)} over the common denominator abc(a-b)(a-c)(b-c) produces numerators a(b-c), -b(a-c) and c(a-b), which add to zero, so the sum equals 0. Similarly, \frac{1}{x(x-y)(x-z)}+\frac{1}{y(y-z)(y-x)}+\frac{1}{z(z-x)(z-y)}=\frac{1}{xyz}, and a quick numerical check with x=1, y=2, z=3 gives \frac12-\frac12+\frac16=\frac16. For three distinct quantities A, B, C, the cyclic sum \frac{A^k}{(A-B)(A-C)}+\frac{B^k}{(B-C)(B-A)}+\frac{C^k}{(C-A)(C-B)} equals 0 for k=0 and k=1, and equals 1 for k=2. This is the Lagrange interpolation pattern, so with A=x^2, B=y^2, C=z^2 and k=2 the expression is the constant 1, independent of x, y and z. Grouping terms usually beats brute-force addition. In \frac{x+1}{x-1}-\frac{x-1}{x+1} the numerator becomes (x+1)^2-(x-1)^2=4x, giving \frac{4x}{x^2-1}. Because x^2-1=-(1-x^2), this is the negative of \frac{4x}{1-x^2}. Similarly, \frac{8x}{1-x^4}-\frac{4x}{1+x^2} becomes \frac{4x}{1-x^2} once 1-x^4=(1-x^2)(1+x^2) is used. Spotting such sign reversals of denominators is the key to sums that collapse to a constant. Numerators of the form x^2-(y+z)^2 are differences of squares, since x^2-(y+z)^2=(x-y-z)(x+y+z), and they should be factorised before cancelling against denominators such as (x+y)^2-z^2=(x+y-z)(x+y+z). Before adding fractions, check that each denominator factors: c^2+bc-a^2-ab=(c-a)(c+a+b), after which the first two terms of a sum combine into \frac{a+b}{c-a}. Ratios whose numerator and denominator are both sums of fractions should be compared directly. Since \frac{x+y}{x-y}=\frac{2x}{x-y}-1, a denominator built from fractions of this type equals 2N-3 plus a constant, where N is the corresponding numerator sum. The constants cancel to give 2N, so the ratio is \frac{N}{2N}=\frac12 whenever N\neq 0. When a condition involves x+\frac1x, divide the numerator and the denominator by x so that the expression depends only on x+\frac1x. For example, \frac{3x}{2x^2+5x+2}=\frac{3}{2\left(x+\frac1x\right)+5}, so x+\frac1x=3 gives \frac{3}{11}. Factor before substituting: \frac{9x^2-1}{3x+1}=3x-1, so the value at x=10 is 29, and the factored form avoids cumbersome arithmetic when x is large. If x^2 is known to equal a multiple of ab, substitution collapses the expression. With x^2=mab, \frac{x^2+ab}{x^2+m^2ab}=\frac{ab(m+1)}{mab(m+1)}=\frac1m, which depends only on the parameter m.

HCF of polynomials

The highest common factor (HCF, also called GCD) of polynomials is the polynomial of highest degree that divides each of them, normally stated up to a constant. The procedure is to factorise every polynomial completely and take each common factor with the smallest power that appears in any of them. When a cubic is involved, test divisors of its constant term to find a root. For example, x^3-19x+30 vanishes at x=2, so it equals (x-2)(x-3)(x+5). The quadratic x^2-5x+6=(x-2)(x-3) then shares exactly the factors (x-2) and (x-3) with it, so the HCF is (x-3)(x-2). Two quadratics that share one linear factor can be tested without full factorisation. If (x-k) divides both x^2+ax+b and x^2+cx+d, then k is a common root, so k^2+ak+b=0 and k^2+ck+d=0. Subtracting gives (a-c)k=d-b, hence k=\frac{d-b}{a-c} whenever a\neq c. For instance, x^2+3x+2 and x^2+5x+6 give k=\frac{6-2}{3-5}=-2, which matches the common factor (x+2). If the common factor is (x+k) and the two quadratics are x^2+px+q and x^2+qx+p with p\neq q, substituting x=-k into both and subtracting gives (p-q)(k+1)=0, so k=-1. The common root is then 1, which requires 1+p+q=0. For two polynomials, LCM \times HCF equals their product, so LCM =\frac{\text{product}}{\text{HCF}}. For example, with (x^2-1) and (x+1)^2, the HCF is (x+1), and the LCM is \frac{(x^2-1)(x+1)^2}{x+1}=(x-1)(x+1)^2. The identity x^4+x^2+1=(x^2+x+1)(x^2-x+1) is a typical case in which a quartic must be split into two quadratics before the HCF or LCM can be read off. Grouping is the main tool when a polynomial has no obvious root. The expression acx^3+bcx^2+adx^2+acdx+bdx+bcd groups as (ax+b)(cx^2+dx+cd), so the linear factor (ax+b) is visible at once. If the leftover quadratics have no common root, that visible linear factor is the HCF. Common factors across several expressions can be read from their factorisations. The factors x^4-13x^2y^2-300y^4=(x^2-25y^2)(x^2+12y^2) come from treating x^2 and y^2 as the variables of a quadratic, while x^3-125y^3=(x-5y)(x^2+5xy+25y^2) comes from the difference of cubes. Only (x-5y) appears in all three, so it is the HCF. Grouping also exposes common factors in higher powers. Here x^8-y^8=(x-y)(x+y)(x^2+y^2)(x^4+y^4), and grouping x^5(x^2+y^2)-y^5(x^2+y^2) shows that the common factor (x-y)(x^2+y^2) expands to x^3-y^3-x^2y+xy^2. For several polynomials the LCM takes every distinct factor with its highest power. Using x^4+x^2y^2+y^4=(x^2+xy+y^2)(x^2-xy+y^2), x^3+y^3=(x+y)(x^2-xy+y^2) and x^3-y^3=(x-y)(x^2+xy+y^2), the LCM is (x+y)(x-y)(x^2+xy+y^2)(x^2-xy+y^2), which equals x^6-y^6. The HCF of two numbers that determine polynomial factors can depend on their arithmetic. If x and y have HCF H, write x=Ha and y=Hb with \gcd(a,b)=1. Then \gcd(x+y,\,x-y)=H\gcd(a+b,\,a-b), and \gcd(a+b,a-b) divides 2, so the common factor H or 2H can occur depending on a and b. A useful identity is a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca). Putting c=-1 factors x^3+y^3+3xy-1 with the linear factor (x+y-1). Since (x+y)^4-1=(x+y-1)(x+y+1)\left((x+y)^2+1\right), the only common factor left is (x+y-1).

Factor theorem

The factor theorem states that (x-a) is a factor of a polynomial f(x) exactly when f(a)=0. It is the fastest test for linear factors, and the candidate values of a are taken from the factors of the constant term. For example, f(1)=0 for x^3-6x^2+11x-6, so (x-1) is a factor and the cubic equals (x-1)(x-2)(x-3). A product form can hide a factor as well. In (x-1)(x-2)(x-4)-90, substituting x=7 gives 6\cdot 5\cdot 3-90=0, so (x-7) divides it. To test whether (x+a) divides a polynomial, evaluate it at x=-a. For x^3+2x^2-x-2 at x=-2 the value is -8+8+2-2=0, so (x+2) is a factor. The same single evaluation decides the question for a quintic, and a sign error in one coefficient changes the value at x=-2 by twice that term, so the arithmetic must be checked carefully. Rational roots give factors of the form (ax-b), and the test is f\left(\frac{b}{a}\right)=0. A polynomial may be divisible by two such factors at once. For instance, f\left(\frac12\right)=0 and f\left(-\frac12\right)=0 for 4x^3+12x^2-x-3, so both (2x-1) and (2x+1) divide it, and indeed 4x^3+12x^2-x-3=(4x^2-1)(x+3). When a divisibility statement uses the word only, each such factor must be checked. If (x-a)^m divides f(x), then a is a root of multiplicity at least m. For a quartic divisible by (x-1)^3, the remaining factor is linear, (x-r), and the constant term fixes r. Since (x-1)^3(x+1)=x^4-2x^3+2x-1 has constant term -1, the other factor is x+1. Grouping can expose repeated roots. The polynomial 1-x-x^n+x^{n+1} equals (1-x)(1-x^n), and 1-x^n=(1-x)(1+x+\cdots+x^{n-1}) for every natural number n, so it is divisible by (1-x)^2=1-2x+x^2. Divisibility by x^2+1 can be checked with the complex root x=i. For x^{4k}+x^{4k+2}+x^{4k+4}+x^{4k+6}, substituting x=i gives 1-1+1-1=0. The polynomial also factorises as x^{4k}(1+x^2)(1+x^4), so it is divisible by x^4+1 as well. A quadratic that splits into linear factors (x-ay)(x-bz) divides a polynomial only if the polynomial vanishes at both x=ay and x=bz. Substituting each value, with y and z treated as constants, gives relations among the coefficients, and these relations fix the constants that the question asks about. When a polynomial identity holds for all x, the factor theorem and coefficient comparison work together. After a common factor on both sides is cancelled, the remaining equation is compared coefficient by coefficient. For example, (x-1)(x+q)=x^2+3x-4 gives q=4 by comparing the coefficient of x, and the constant term -q=-4 confirms it.

Sum of cubes identity

The sum and difference of two cubes factorise into a linear factor times a quadratic factor. For any real numbers a and b:

• a^3+b^3=(a+b)(a^2-ab+b^2)
• a^3-b^3=(a-b)(a^2+ab+b^2)

The sign in the quadratic factor is the key point. The sum has -ab in its quadratic factor, and the difference has +ab. Confusing the two is a common error in options that offer both forms.

The cube sum can also be written without factoring: a^3+b^3=(a+b)^3-3ab(a+b). This form is useful when a+b and ab are known, because the cube sum can then be found without knowing a and b separately. For example, if a+b=3 and ab=2, then a^3+b^3=27-18=9. The same method works with reciprocals. If \frac{1}{a}+\frac{1}{b} is given, then ab is found first from \frac{a+b}{ab}.

A quotient whose denominator is the quadratic factor of a difference of cubes simplifies to the linear factor. Decimal numbers are often disguised in this way: a value such as 0.064 is 0.4^3, and 2.16 is 5.4\times0.4, so the expression matches \frac{a^3-b^3}{a^2+ab+b^2}=a-b.

The three-variable form is a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca). Expressions that combine (x+y+z)^3 and xyz, where x,y,z are sums of pairs of a,b,c, reduce to a multiple of a^3+b^3+c^3. A quick check is to put a=b=c=1 and compare the values.

When the numerator and the denominator both contain the same symmetric product, the ratio is identically 1. The method is to write each side as the same product (a+b+c)(a^2+b^2+c^2-ab-bc-ca) and cancel.

Some cubic expressions contain surd coefficients. Substituting A=\sqrt{3}x, B=\sqrt{2}y, C=\sqrt{6} turns the expression into A^3+B^3+C^3-3ABC, and the second factor is A^2+B^2+C^2-AB-BC-CA. Recognising this form lets you pick the correct factor.

Factors such as 1-p^3 and 1+p^3 expand as (1-p)(1+p+p^2) and (1+p)(1-p+p^2). Their difference is -2p^3, whose sign depends on p. A factorised form therefore does not by itself decide which of two quantities is larger.

For reciprocal sums, put u=\frac{1}{a} and v=\frac{1}{b}. Then uv=\frac{1}{2}\left[(u+v)^2-(u^2+v^2)\right], and u^3+v^3=(u+v)^3-3uv(u+v). This two-step method is the standard way to find cube sums of reciprocals.

Symmetric sum of squares identity

The symmetric expression a^2+b^2+c^2-ab-bc-ca is central to this topic. It can be written as $a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}\left[(a-b)^2+(b-c)^2+(c-a)^2\right]$ This shows that the expression is never negative. It is zero exactly when a=b=c, and it is strictly positive whenever a, b, c are distinct.

Because a sum of non-negative squares is zero only when each square is zero, a statement that gives the sum of squared differences equal to zero forces equality. The same reasoning applies to x^2+y^2+z^2-xy-yz-zx=0, which forces x=y=z. By contrast, x^3+y^3+z^3-3xyz=0 factorises as (x+y+z)(x^2+y^2+z^2-xy-yz-zx)=0, so x+y+z=0 is also possible without equality. Checking one statement alone against this factorisation shows which one is sufficient.

Two useful forms for working with the sum of squares are a^2+b^2=(a+b)^2-2ab and a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca). For example, if a product is 12 and the sum of squares is 25, then (a+b)^2=25+24=49, so a+b=7 (taking the positive value). These forms are used in problems that give a product and sum of squares and ask for a mean or a sum.

The identity a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca) can be applied with a=x^2, b=y^2, c=z^2. This turns an expression in sixth powers into a product involving x^2+y^2+z^2 and a symmetric quadratic in p,q,r.

Some expressions built from this identity collapse to a constant. When a result seems to depend on variables, test with simple values such as x=1 and y=z=0, or with a set of values that satisfies the given relation. If the expression gives the same number for several sets of values, that number is the identical value.

If 2s=a+b+c, then the expression s^2+(s-a)(s-b)+(s-b)(s-c)+(s-c)(s-a) reduces by expansion. Using a+b+c=2s in the linear terms gives the result ab+bc+ca. Putting a=b=c=1 gives 3 on both sides, which confirms the result.

If a sum of two non-negative quantities is zero, each quantity must be zero. For example, (a+b+c)^2-3(ab+bc+ca) is non-negative, and 3(x^2+y^2+z^2) is also non-negative. Their sum is zero only when a=b=c and x=y=z=0.

For a triangle with sides a, b, c, the condition a^2+b^2+c^2=ab+bc+ca means a=b=c, since the expression equals half the sum of squared differences. Another form, (a-b)^2+2(a-c)^2+2(b-c)^2=0, also forces a=b=c. Either condition alone is therefore enough to conclude that the triangle is equilateral.

When the sum of squared differences and the sum of squares are known, the product sum follows directly. From (a-b)^2+(b-c)^2+(c-a)^2=2(a^2+b^2+c^2)-2(ab+bc+ca), the value of ab+bc+ca is found, and then (a+b+c)^2 follows. This gives the value of a+b+c up to sign, so both signs must be given.

Perfect square identities

The basic perfect square identities are (a+b)^2=a^2+2ab+b^2 and (a-b)^2=a^2-2ab+b^2. A quadratic expression is a perfect square when it matches one of these forms. These identities are used both to factorise and to test whether an expression can be written as a square.

A quadratic Ax^2+Bx+C is a perfect square for every value of x exactly when its discriminant is zero, that is B^2-4AC=0, and A is a positive perfect square. Equivalently, B^2=4AC, and the expression equals \left(\sqrt{A}x\pm\sqrt{C}\right)^2 with the appropriate sign of \sqrt{C}. The condition gives an equation in the parameter, which may have two roots, and each root must be checked.