Factoring by Substitution
Generally, factoring polynomials means separating a polynomial into its component polynomials. Sometimes, when polynomials are particularly complicated, it is easiest to substitute a simple term and factor down. Sometimes this looks like the example below, where a multi-variable polynomial is substituted for a single variable, \(S\) in this case, which stands in for the single polynomial until the overall polynomial is factored down.
This is a quick technique to use when faced with complicated polynomials, particularly if subsets of that polynomial repeat. For instance, in the example below, \(x-y\) repeats twice in the equation. Sometimes manipulation of the overall polynomial is necessary to see this repetition. It is also useful when faced with complicated components like roots, imaginary numbers, and other non-real, non-integer numbers.
Factor \( (x - y)(x - y - 1) - 20 \).
Let \( S = x - y \). Then substituting \( S \) in for \( x - y\) for ease of computation, we get
\[ \begin{align} (x - y)(x - y - 1) - 20 &= (S)(S - 1) - 20 \\ &= S^2 - S - 20 \\ &= (S - 5)(S + 4) \\ &= (x - y - 5)(x - y + 4). \ _\square \end{align} \]
Factor \( 3(4x + 5)^2 - 2(4x + 5) - 1 \).
Let \( S \) = \( 4x + 5 \). Then substituting \( S \) in for \( 4x+ 5 \), we get
\[ \begin{align} 3(4x + 5)^2 - 2(4x + 5) - 1 &= 3S^2 - 2S - 1 \\ &= (3S + 1)(S - 1) \\ &= (3(4x+5) + 1)(4x + 5 -1) \\ &= (12x + 15 + 1)(4x + 5 - 1) \\ &= (12x + 16)(4x + 4) \\ &= 4(3x + 4) \cdot 4(x+1) \\ &= 16(3x + 4)(x + 1). \ _\square \end{align} \]
Challenging Example Questions
The following is an example of substitution of imaginaries:
Solve for all roots of the polynomial
\[x^4-5ix^3+19x^2-125ix-150.\]
We begin by substituting \(a=-ix\) (be careful of +/- signs), which leaves us with
\[a^4-5a^3-19a^2+125a-150.\]
Looks much more factorable, doesn't it? Now it's just grouping, grouping, and more grouping. (If you want, you can use this trick.)
\[\begin{align} a^4-5a^3-19a^2+125a-150 &=a^3(a-5)-(19a^2-125a+150)\\ &=a^3(a-5)-(19a-30)(a-5)\\ &=(a^3-19a+30)(a-5). \end{align}\]
Using some logic, since there's a gap in powers (\(a^3\) followed by \(a\)), there should be a difference of squares. By remainder theorem, since 5 is a solution, we can check -5.
\[-125+95+30=0.\]
Some synthetic division yields
\[(a^2+5a+6)(a-5)(a+5)=(a+2)(a+3)(a-5)(a+5).\]
Re-substituting \(ix=a\), we arrive at
\[(ix+2)(ix+3)(ix-5)(ix+5). \ _\square\]
Substitution can also be used to minimize and maximize functions.
Find the minimum value of \(4x^2+20xy+30x+25y^2+75y+36\).
The given expression can be rewritten as \((2x+5y)^2+15(2x+5y)+36\). Substituting \(a=2x+5y\), this is equal to \(a^2+15a+36\). Completing the square gives \[\begin{align} a^2+15a+36 &=\left(a+\frac{15}{2}\right)^2+36-\left(\frac{15}{2}\right)^2\\ &=\left(a+\frac{15}{2}\right)^2-\frac{81}{4}, \end{align}\] which implies that the minimum value of the given expression is \(-\frac{81}{4}.\) \(_\square\)
Substitution can also be used for irrational coefficients.
Find the sum of all positive roots of \(18\sqrt 3x^3+51x^2-31\sqrt 3x-12\).
Substituting \(a=\sqrt{3}x\) gives
\[\begin{align} 6a^3+17a^2-31a-12 &=(x+4)(6a^2-7a-3)\\ &=(x+4)(3a+1)(2a-3). \end{align}\]
Since we are interested in positive roots, \(a=\dfrac{3}{2},\) which implies that the sum of all positive roots of the given expression is \[x=\frac{a}{\sqrt{3}}=\frac{\sqrt3}{2}. \ _\square\]