Transforming the Roots of a Polynomial

Double every root without finding one.

New roots from old

The equation x² − 5x + 6 = 0 has roots 2 and 3. Suppose an equation is wanted whose roots are each of these doubled, 4 and 6. Other questions ask for the roots shifted, such as α + 3, or turned into reciprocals, 1/α, or squared, α².

For a quadratic with easy roots, the answer could be built from the new roots directly. The method here never finds the old roots at all, so it works just as well for a cubic whose roots are hard to find.

Write the transformation, then reverse it

Call a new root y and an old root x. Each new root is twice an old one, so y = 2x. Rearrange to give the old in terms of the new: x = y/2.

That reversed form is what goes into the equation. The old equation is true when x is an old root, and x = y/2 is an old root exactly when y is twice one.

Substitute and tidy up

Put y/2 in place of every x in x² − 5x + 6 = 0: (y/2)² − 5(y/2) + 6 = 0, which is y²/4 − 5y/2 + 6 = 0.

Multiply every term by 4 to clear the fractions: y² − 10y + 24 = 0. Multiplying through by 4 does not change which values of y make it zero, so the roots are unchanged.

Check by factoring

y² − 10y + 24 = (y − 4)(y − 6), so the roots are 4 and 6, exactly the doubled pair.

The sum and product agree too. The old roots add to 5 and multiply to 6. Doubled, they add to 10 and multiply to 4 × 6 = 24, and y² − 10y + 24 has sum −b/a = 10 and product c/a = 24.

xy

The gold curve is y = x² − 5x + 6, crossing the x-axis at 2 and 3. The plain curve is the same expression with x/2 in place of x, which is ¼(x² − 10x + 24). It is the gold curve stretched away from the y-axis by a factor of 2, so its crossings are at 4 and 6.

A shift: roots α + 3

For roots 3 more than the old ones, y = x + 3, so x = y − 3. Substitute: (y − 3)² − 5(y − 3) + 6 = y² − 6y + 9 − 5y + 15 + 6 = y² − 11y + 30.

The equation is y² − 11y + 30 = 0, and y² − 11y + 30 = (y − 5)(y − 6), so its roots are 5 and 6: the old roots 2 and 3, each moved up by 3. On the graph, putting x − 3 in place of x moves the curve 3 to the right.

xy

The gold curve is y = x² − 5x + 6 again. The plain curve is the same expression with x − 3 in place of x, which is x² − 11x + 30. It is the gold curve moved 3 to the right, so its crossings move from 2 and 3 to 5 and 6.

Reciprocals and squares

For roots 1/α, y = 1/x, so x = 1/y. Substitute: 1/y² − 5/y + 6 = 0. Multiply through by y²: 1 − 5y + 6y² = 0, which is 6y² − 5y + 1 = 0. It factors as (2y − 1)(3y − 1) = 0, so the roots are ½ and ⅓, the reciprocals of 2 and 3. The coefficients have come out in reverse order.

For roots α², y = x², so x = √y. Substitute: y − 5√y + 6 = 0. Move the root to one side, y + 6 = 5√y, and square both sides: y² + 12y + 36 = 25y. So y² − 13y + 36 = 0, which factors as (y − 4)(y − 9) = 0, with roots 4 and 9, the squares of 2 and 3.

A cubic

The method is the same for any degree. x³ − 6x² + 11x − 6 = 0 has roots 1, 2 and 3. For the roots doubled, put x = y/2: y³/8 − 6y²/4 + 11y/2 − 6 = 0. Multiply by 8: y³ − 12y² + 44y − 48 = 0.

Check with the three sums of its roots, 2, 4 and 6. They add to 12, the products in pairs give 8 + 24 + 12 = 44, and the product is 48, which match −b/a, c/a and −d/a for y³ − 12y² + 44y − 48.

The usual mistakes

Substituting the transformation instead of its reverse. Putting x = 2y into x² − 5x + 6 gives 4y² − 10y + 6 = 0, which is 2y² − 5y + 3 = (2y − 3)(y − 1) = 0, with roots 1 and 3/2: the old roots halved, not doubled.

Shifting the wrong way. For roots α + 3, x = y − 3 goes in. Putting x = y + 3 gives roots 3 less than the old ones, −1 and 0.

Changing only some coefficients. Doubling the constant to get y² − 5y + 12, or doubling the sum to get y² − 10y + 6, does not double both roots: doubling scales the sum by 2 and the product by 4.

Scaling every coefficient by the same number. 2x² − 10x + 12 = 0 has the same roots as x² − 5x + 6 = 0, because dividing by 2 gives the old equation back.

A sensor on a fairground ride

In the application below, the moments a cart passes a sensor are the roots of a cubic. A second clock reads 2 seconds more, so z = x + 2 and x = z − 2 goes in. On a faster setting every time is halved, so y = x/2 and x = 2y goes in.

Worked example: A Sensor on a Fairground Ride: The Equation for the Times on a Second Clock and on a Faster Setting

Question A sensor on a fairground ride records the moments a cart passes it. The times are the roots of x3 − 9x2 + 23x − 15 = 0, where x is the number of seconds after the cart leaves the station. (a) A second clock starts when the safety bar locks, 2 seconds before the cart leaves. Without solving the cubic, write the equation whose roots are the times on the second clock. (b) On a faster setting every time is halved. Write the equation for the new times with whole-number coefficients, and use its coefficients to find the sum of the three new times. Check both equations by solving the first cubic.

  1. 1.A time z on the second clock is 2 seconds more than the time x: z = x + 2, so x = z − 2. Substitute into the cubic: (z − 2)3 − 9(z − 2)2 + 23(z − 2) − 15 = 0.

    −16−80816012345678secondsvalue of the cubicxz = x + 2, so x = z − 2(z − 2)3− 9(z − 2)2+ 23(z − 2) − 15 = 0
    −16−80816012345678secondsvalue of the cubicxz = x + 2, so x = z − 2(z − 2)3− 9(z − 2)2+ 23(z − 2) − 15 = 0
    A time on the second clock is z = x + 2, so x = z − 2 goes into the cubic: (z − 2)3 − 9(z − 2)2 + 23(z − 2) − 15 = 0.

  2. 2.Expand each bracket: z3 − 6z2 + 12z − 8 − 9z2 + 36z − 36 + 23z − 46 − 15 = 0. (a) Collect the terms: z3 − 15z2 + 71z − 105 = 0.

    −16−80816012345678secondsvalue of the cubicxzz3− 6z2+ 12z − 8 − 9z2+ 36z − 36+ 23z − 46 − 15 = 0z3− 15z2+ 71z − 105 = 0
    −16−80816012345678secondsvalue of the cubicxzz3− 6z2+ 12z − 8 − 9z2+ 36z − 36+ 23z − 46 − 15 = 0z3− 15z2+ 71z − 105 = 0
    (a) Expanding and collecting gives z3 − 15z2 + 71z − 105 = 0. Its curve is the first one moved 2 to the right.

  3. 3.On the faster setting a new time is y = x2, so x = 2y. Substitute: 8y3 − 36y2 + 46y − 15 = 0, which already has whole-number coefficients.

    −16−80816012345678secondsvalue of the cubicxyy = x/2, so x = 2y8y3− 36y2+ 46y − 15 = 0
    −16−80816012345678secondsvalue of the cubicxyy = x/2, so x = 2y8y3− 36y2+ 46y − 15 = 0
    On the faster setting x = 2y, which gives 8y3 − 36y2 + 46y − 15 = 0. Its curve is the first one squeezed to half the width.

  4. 4.(b) The equation is 8y3 − 36y2 + 46y − 15 = 0, and the sum of its roots is 368 = 4.5 seconds, half of the 9 seconds from the first cubic.

    −16−80816012345678secondsvalue of the cubicxysum of the roots = 36/8 = 4.5 shalf of the 9 s from the first cubic
    −16−80816012345678secondsvalue of the cubicxysum of the roots = 36/8 = 4.5 shalf of the 9 s from the first cubic
    (b) The sum of the new times is 368 = 4.5 seconds, half of the 9 seconds from the first cubic.

  5. 5.Check: x = 1 gives 1 − 9 + 23 − 15 = 0, and dividing out (x − 1) leaves x2 − 8x + 15 = (x − 3)(x − 5). The cart passes at 1, 3 and 5 seconds; on the second clock at 3, 5 and 7, which add up to 15 and multiply to 105; on the faster setting at 0.5, 1.5 and 2.5, which add up to 4.5.

    −16−80816012345678secondsvalue of the cubicxzyx: 1, 3 and 5 sz: 3, 5 and 7 sy: 0.5, 1.5 and 2.5 s
    −16−80816012345678secondsvalue of the cubicxzyx: 1, 3 and 5 sz: 3, 5 and 7 sy: 0.5, 1.5 and 2.5 s
    The first cubic factorizes as (x − 1)(x − 3)(x − 5), so the curves cross the axis at 1, 3, 5, at 3, 5, 7 and at 0.5, 1.5, 2.5.

Answer: (a) z3 − 15z2 + 71z − 105 = 0; (b) 8y3 − 36y2 + 46y − 15 = 0, whose roots add up to 4.5 seconds; the times are 1, 3 and 5 seconds, then 3, 5 and 7 seconds, then 0.5, 1.5 and 2.5 seconds

Common mistakes

  • Putting z + 2 in place of x. That gives an equation whose roots are 2 less than the old ones, the times −1, 1 and 3. The substitution is the old unknown in terms of the new one: x = z − 2.
  • Halving the coefficients, or putting y2 in place of x. A new time y is half an old time, so x = 2y, and it is 2y that goes into the cubic.

More polynomials and the binomial theorem problems, worked step by step →

Practice Transforming the Roots of a Polynomial in the app