P1.2.6 - Algebraic manipulation of polynomials

P1.2.6 - Algebraic manipulation of polynomials

Polynomial manipulation is the art of changing the form of an expression without changing its value. In this lesson you will expand, collect, factorise, divide by linear expressions, use the factor theorem, and simplify rational expressions. The thread running through all of it is structure: look for the factor, root, or common expression that makes the algebra smaller.

See the structure before expanding

Polynomial

A polynomial in x is a finite sum of terms such as ax^n, where the powers of x are non-negative integers and the coefficients are constants.

For example, 3x^4 - 2x + 7 is a polynomial, but x^{-1} + 4 and sqrt(x) + 2 are not polynomials in this sense.

Algebraic manipulation usually means choosing between two opposite moves:

  • expand brackets and collect like terms when a single polynomial form is useful
  • factorise when a product form reveals common factors or roots

Worked example: simplify

P(x)=(x−2)(x+5)+3x(x−2).P(x)=(x-2)(x+5)+3x(x-2).

You could expand everything, but the common factor (x-2) is visible:

P(x)=(x−2)((x+5)+3x)=(x−2)(4x+5).\begin{aligned} P(x)&=(x-2)\big((x+5)+3x\big)\\ &=(x-2)(4x+5). \end{aligned}

If an expanded form is required, multiply out:

(x−2)(4x+5)=4x2−3x−10.(x-2)(4x+5)=4x^2-3x-10.

Both forms are equivalent. The factorised form is better for seeing zeros; the expanded form is better for collecting terms or comparing coefficients.

See the structure before expanding Continued

The sign before a bracket belongs to every term in that bracket. Many polynomial errors are not "hard maths" errors; they are one missed sign.

Divide by a linear expression

In this specification, polynomial division means division by a linear expression such as (x-3), (x+2), (2x-1), or (3x+4). When a polynomial is divided by a linear expression, the result has the form

f(x)=(linear divisor)(quotient)+constant remainder.f(x)=(\text{linear divisor})(\text{quotient})+\text{constant remainder}.

If the constant remainder is zero, the divisor is a factor.

Worked example: divide

2x3−3x2−11x+62x^3-3x^2-11x+6

by (x-3).

Start by matching the leading term. Since 2x^3 / x = 2x^2, the first term of the quotient is 2x^2:

2x3−3x2−11x+6−(2x3−6x2)=3x2−11x+6.\begin{aligned} 2x^3-3x^2-11x+6 &-(2x^3-6x^2)\\ &=3x^2-11x+6. \end{aligned}

Now 3x^2 / x = 3x, so subtract 3x(x-3)=3x^2-9x:

3x2−11x+6−(3x2−9x)=−2x+6.3x^2-11x+6-(3x^2-9x)=-2x+6.

Finally, -2x / x = -2, and subtract -2(x-3)=-2x+6:

−2x+6−(−2x+6)=0.-2x+6-(-2x+6)=0.

So

2x3−3x2−11x+6=(x−3)(2x2+3x−2).2x^3-3x^2-11x+6=(x-3)(2x^2+3x-2).

The quotient is 2x^2+3x-2 and the remainder is 0.

Divide by a linear expression Continued

If a power is missing, write its coefficient as zero before dividing. For example, treat x^3-7x+6 as x^3+0x^2-7x+6.

Use the factor theorem

The factor theorem links substitution to factorisation.

Factor theorem

For a polynomial f(x), if f(r)=0, then (x-r) is a factor of f(x).

This is really a zero-remainder statement. Dividing by (x-r) gives a constant remainder; if substituting x=r gives zero, that remainder is zero.

For a factor such as (ax-b), first find the value of x that makes the factor zero:

ax−b=0⇒x=ba.ax-b=0 \quad \Rightarrow \quad x=\frac{b}{a}.

So if f(b/a)=0, then (ax-b) is a factor of f(x). For example, to test whether (3x-2) is a factor, substitute x=2/3, not x=2.

Worked example: given

g(x)=2x3−9x2+(k+5)x+2k,g(x)=2x^3-9x^2+(k+5)x+2k,

and given that (x-2) is a factor of g(x), find k.

Since (x-2) is a factor, the factor theorem gives g(2)=0:

g(2)=2(2)3−9(2)2+(k+5)(2)+2k=16−36+2k+10+2k=4k−10.\begin{aligned} g(2)&=2(2)^3-9(2)^2+(k+5)(2)+2k\\ &=16-36+2k+10+2k\\ &=4k-10. \end{aligned}

Set this equal to zero:

4k−10=0⇒k=52.4k-10=0 \quad \Rightarrow \quad k=\frac{5}{2}.

Explain It Back

Use this as a self-explanation check after the section above. It is for diagnosing what you can already explain, not for learning new material from scratch.

Notice the sign: (x+1) means the root is -1. That one sign is a very common source of lost marks.

Factorise cubics using one root

For a cubic, the factor theorem usually gives the first linear factor. Then algebraic division reduces the cubic to a quadratic, which you can factorise by ordinary quadratic methods.

Worked example: factorise

f(x)=x3+3x2−4.f(x)=x^3+3x^2-4.

Try small integer values. Since

f(1)=1+3−4=0,f(1)=1+3-4=0,

(x-1) is a factor. Divide x^3+3x^2+0x-4 by (x-1):

x3+3x2+0x−4=(x−1)(x2+4x+4).x^3+3x^2+0x-4=(x-1)(x^2+4x+4).

Now factorise the quadratic:

x2+4x+4=(x+2)2.x^2+4x+4=(x+2)^2.

Therefore

x3+3x2−4=(x−1)(x+2)2.\boxed{x^3+3x^2-4=(x-1)(x+2)^2}.

Here is a second example with fractional roots hidden inside linear factors:

p(x)=6x3+11x2−x−6.p(x)=6x^3+11x^2-x-6.

Testing x=-1 gives

p(−1)=−6+11+1−6=0,p(-1)=-6+11+1-6=0,

so (x+1) is a factor. Dividing gives

6x3+11x2−x−6=(x+1)(6x2+5x−6).6x^3+11x^2-x-6=(x+1)(6x^2+5x-6).

Now

6x2+5x−6=(2x+3)(3x−2),6x^2+5x-6=(2x+3)(3x-2),

so

6x3+11x2−x−6=(x+1)(2x+3)(3x−2).\boxed{6x^3+11x^2-x-6=(x+1)(2x+3)(3x-2)}.

The factor (3x-2) corresponds to the root x=2/3. Edexcel can expect you to recognise factors of the form (ax-b), not only (x-r).

Factorise cubics using one root Continued

If the quadratic factor does not factorise neatly, leave it as an irreducible quadratic unless the question asks you to solve the equation. The instruction "factorise fully" usually expects all available rational linear factors.

Simplify rational expressions safely

A rational expression is a fraction whose numerator and denominator are polynomials. To simplify it, factorise first, then cancel common factors.

Worked example: simplify

x2+5x+6x2−x−6.\frac{x^2+5x+6}{x^2-x-6}.

Factorise the numerator and denominator:

x2+5x+6x2−x−6=(x+2)(x+3)(x+2)(x−3).\frac{x^2+5x+6}{x^2-x-6} =\frac{(x+2)(x+3)}{(x+2)(x-3)}.

Now cancel the common factor (x+2):

(x+2)(x+3)(x+2)(x−3)=x+3x−3.\frac{(x+2)(x+3)}{(x+2)(x-3)} =\frac{x+3}{x-3}.

The original denominator was zero when x=-2 or x=3, so the simplified expression is equivalent to the original only for

x≠−2,x≠3.x\ne -2,\quad x\ne 3.

Cancel common factors, not common terms. You may cancel (x+2) from a product, but you may not cancel the x from something like (x+2)/(x+5).

Sometimes a rational expression is top-heavy, meaning the numerator has degree at least as large as the denominator. Then algebraic division may be the cleanest form.

Worked example: simplify

x2+5x+7x+2.\frac{x^2+5x+7}{x+2}.

Divide the numerator by x+2:

x2+5x+7=(x+2)(x+3)+1.x^2+5x+7=(x+2)(x+3)+1.

So

x2+5x+7x+2=x+3+1x+2,x≠−2.\frac{x^2+5x+7}{x+2} =x+3+\frac{1}{x+2}, \qquad x\ne -2.

This is not partial fractions; it is just linear algebraic division to rewrite an improper rational expression.

Simplify rational expressions safely Continued

That final restriction matters even though (x-2) has disappeared from the simplified expression. Simplifying changes the form, not the original set of allowed inputs.