1.1 - Shifting and scaling roots | Cambridge A-Level Further Mathematics

You can form an equation for shifted or scaled roots without finding the original roots. You will choose the substitution in the correct direction, expand it carefully and check the transformed equation using its roots and coefficients.

Choosing the substitution

If every root increases by $2$ , why does the substitution contain a minus sign? The original polynomial needs the old root as its input, so you must undo the change before substituting.

Suppose an old root is $x=4$ , and the new root is $y=x+2=6$ . To feed $4$ into the old polynomial when $y=6$ , use $x=y-2$ . Substituting $y+2$ would feed in $8$ , a different number.

Linear root transformation

A linear root transformation sends each old root $r$ to $kr+h$ , where $k,h$ are constants and $k\ne0$ . To find its equation, set

y=kx+h\qquad\text{and hence}\qquad x=\frac{y-h}{k}.

Substitute the expression for $x$ into the original equation.

For $f(x)=0$ , the new equation is $f((y-h)/k)=0$ . Each old root produces a new root, and the inverse substitution sends every new root back to an old one. This explains why no root is lost or added when $k\ne0$ .

Worked example 1: undoing a shift

Question. The roots of $x^2-3x-4=0$ are $\alpha,\beta$ . Write an equation, without expanding, whose roots are $\alpha+2,\beta+2$ .

Plan. Name a new root $y=x+2$ , rearrange for $x$ , then replace every occurrence of $x$ .

\begin{aligned} y&=x+2\quad\Rightarrow\quad x=y-2 \\ 0&=(y-2)^2-3(y-2)-4. \end{aligned}

Answer. $(y-2)^2-3(y-2)-4=0$ .

Check. The old equation factorises as $(x-4)(x+1)=0$ . The proposed new roots $6,1$ make $y-2$ equal to $4,-1$ , respectively, so both satisfy the new equation.

Add a scale factor while keeping the same shift and original equation. Undo the shift first, then divide by the scale.

Worked example 2: undoing a scale and a shift

Question. Write an unexpanded equation whose roots are $2\alpha+2,2\beta+2$ , for the same original equation $x^2-3x-4=0$ .

Plan. Solve $y=2x+2$ for the old variable before substituting.

\begin{aligned} x&=\frac{y-2}{2} \\ 0&=\left(\frac{y-2}{2}\right)^2-3\left(\frac{y-2}{2}\right)-4. \end{aligned}

Answer. $\left(\frac{y-2}{2}\right)^2-3\left(\frac{y-2}{2}\right)-4=0$ .

Check. The old roots $4,-1$ become $10,0$ . Substituting either value makes the old-variable expression $(y-2)/2$ equal to its original root.

These examples stop before expansion so the direction of the substitution remains visible. The next parts carry it through to a polynomial equation.

Watch out

For new roots $k\alpha+h$ , substituting $kx+h$ into the old polynomial reverses the intended change. Write “new = change applied to old”, then rearrange for “old”.

Choose and apply the inverse change in a new equation.

Shifting roots

To increase every root by $h$ , substitute $x=y-h$ . Brackets matter: every power and every linear term must use the same replacement.

Equation for shifted roots

If $f(x)=0$ has roots $\alpha,\beta,\ldots$ , then

f(y-h)=0

has roots $\alpha+h,\beta+h,\ldots$ , with the same multiplicities. The constant $h$ may be positive, negative or zero.

Not in the formula booklet: learn it.

Expand one power at a time before collecting equal powers. This makes a sign error easier to locate than a single long line of arithmetic.

Worked example 3: shifting all roots upwards

Question. The roots of $x^3-x^2-4x+4=0$ are $\alpha,\beta,\gamma$ . Find a cubic equation with roots $\alpha+2,\beta+2,\gamma+2$ .

Plan. Substitute $x=y-2$ and collect powers of $y$ .

\begin{aligned} 0&=(y-2)^3-(y-2)^2-4(y-2)+4 \\ &=y^3-6y^2+12y-8 \\ &\quad-y^2+4y-4-4y+8+4 \\ &=y^3-7y^2+12y. \end{aligned}

Answer. $y^3-7y^2+12y=0$ .

Check. The old roots are $1,2,-2$ , from $(x-1)(x^2-4)$ . Their shifted values are $3,4,0$ , and the new polynomial is $y(y-3)(y-4)$ .

Change only the direction of the shift. Subtracting $2$ from the roots now requires adding $2$ inside the original polynomial.

Worked example 4: shifting all roots downwards

Question. For the same original equation, find a cubic equation with roots $\alpha-2,\beta-2,\gamma-2$ .

Plan. From $y=x-2$ , obtain $x=y+2$ , then expand.

\begin{aligned} 0&=(y+2)^3-(y+2)^2-4(y+2)+4 \\ &=y^3+6y^2+12y+8 \\ &\quad-y^2-4y-4-4y-8+4 \\ &=y^3+5y^2+4y. \end{aligned}

Answer. $y^3+5y^2+4y=0$ .

Check. The new roots are $-1,0,-4$ . Their equation is $y(y+1)(y+4)=0$ , confirming both the sign and the zero constant.

A shift can create a zero root. That happens here because one old root is exactly the amount subtracted; it causes no difficulty for a linear transformation.

Problem-solving

For an equation of degree $n$ , adding $h$ to every root increases the root sum by $nh$ . Use this to check the coefficient of the next-to-highest power after expansion.

Work through a different shift without finding the original roots.

Scaling roots

Scaling roots means changing the input variable, rather than multiplying the whole equation by a constant. Multiplying $f(x)=0$ by $4$ leaves its roots unchanged; replacing $x$ by $y/4$ makes the new roots four times as large.

For the combined change $y=kx+h$ , substitute $x=(y-h)/k$ . Multiplying afterwards by a nonzero constant can clear fractions without changing the new roots. For a degree- $n$ equation, multiplying by $k^n$ always clears the denominators introduced by this substitution, although a smaller multiplier sometimes suffices.

Worked example 5: doubling the roots

Question. The roots of $2x^3+3x^2-8x+3=0$ are $\alpha,\beta,\gamma$ . Find a cubic equation with roots $2\alpha,2\beta,2\gamma$ .

Plan. Substitute $x=y/2$ , then clear the fractions.

\begin{aligned} 0&=2\left(\frac y2\right)^3+3\left(\frac y2\right)^2-8\left(\frac y2\right)+3 \\ &=\frac{y^3}{4}+\frac{3y^2}{4}-4y+3 \\ 0&=y^3+3y^2-16y+12 &&\text{multiply by }4. \end{aligned}

Answer. $y^3+3y^2-16y+12=0$ .

Check. The original factorisation is $(x-1)(2x-1)(x+3)$ , giving roots $1,\tfrac12,-3$ . Their doubles $2,1,-6$ give $(y-2)(y-1)(y+6)=y^3+3y^2-16y+12$ .

Keep the doubling but add a shift of $1$ . The expression $y-1$ now occupies every position previously occupied by $y$ .

Worked example 6: doubling and then shifting

Question. Find a cubic equation whose roots are $2\alpha+1,2\beta+1,2\gamma+1$ , for the same original equation.

Plan. Substitute $x=(y-1)/2$ , multiply by $4$ , then expand.

\begin{aligned} 0&=(y-1)^3+3(y-1)^2-16(y-1)+12 \\ &=y^3-3y^2+3y-1 \\ &\quad+3y^2-6y+3-16y+16+12 \\ &=y^3-19y+30. \end{aligned}

Answer. $y^3-19y+30=0$ .

Check. The transformed roots are $3,2,-5$ . Their sum is zero, their pair products add to $6-15-10=-19$ , and their product is $-30$ , matching all three coefficients.

The opening minus sign is now accounted for: the new equation undoes the change before using the original polynomial. If the question wants its final equation in $x$ , you may rename the new variable after the calculation, making clear that it now represents the transformed roots.

Watch out

The condition $k\ne0$ matters. When $k=0$ , all transformed values equal $h$ and the inverse substitution does not exist. The method above applies only to nonzero scales, including negative scales.

Finish with a negative scale, which reverses the direction as well as changing the size.

Quick Recap

Write the new root in terms of the old root first.
Rearrange for the old variable before substituting.
A shift by $h$ uses $f(y-h)=0$ .
A nonzero scale and shift use $f((y-h)/k)=0$ ; clear fractions with a nonzero multiplier.

Increasing roots requires a minus sign inside the old polynomial because the substitution must recover the original input.

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1.1 - Shifting and scaling roots

Choosing the substitution

Shifting roots

Scaling roots

Quick Recap

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