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Finding Quadratics
     from Their Zeroes (page 1 of 2)

Sections: Finding quadratics from their zeroes, Finding polynomials from a list of values

Suppose I have some polynomial and I have to solve it. First I have to factor, and then I have to solve each factor. For example:

• Find the zeroes of y = x⁴ + x³ – 7x² – x + 6

I apply the Rational Roots Test, find some zeroes using synthetic division, and finally end up with:

(x + 3)(x – 2)(x + 1)(x – 1) = 0

Solving, I get that x = ±1, –3, 2

How did I get that solution? By solving the factors. Then if I work backwards from the solutions, I can see that, since x = 2 is a solution, for instance, then I must have solved x – 2 = 0, so then x – 2 must have been a factor.

Note that the factors you find from the zeroes are always of the form "(variable) minus (the given zero)". So "(variable) minus (value)" means "(variable) equals (value)" is a solution; that is, if "x – a" is a factor, then "x = a" is a solution, and vice verse. You use this fact to find quadratics from their roots.

• Find a quadratic with zeroes at 4 and –5.

If the zeroes are at x = 4 and at x = –5, then, subtracting, the factors were x – 4 and x – (–5) = x + 5. A quadratic is going to have two linear factors, so these must be the two. Then the quadratic was something like:

(x – 4)(x + 5) = x² – 4x + 5x – 20 = x² + x – 20

Why did I say that the quadratic would be "something like" x² + x – 20? Because they may have divided something out when they solved the original quadratic. For instance, if I had to solve 2x² – 2 = 0, I would first divide off the 2 to get x² – 1 = (x + 1)(x – 1) = 0, with solutions at x = ±1. But if I multiply the factors back, I get (x – 1)(x + 1) = x² – 1, which is not quite what I started with. Any number of other quadratics would also share those same two zeroes, including 5x² – 5, 17x² – 17, and 4 – 4x². If they only give me the zeroes, I can't tell if they divided anything out or not. The quadratic answer I gave in the problem above is good enough, though, because they only asked for "a" quadratic with the given zeroes, not "the" quadratic. But that distinction, between "a" and "the", can be very important.

• Find the general form of the family of quadratics with zeroes at x = 1 and x = 3.

Subtracting, the factors were x – 1 and x – 3. However, I can't tell if they multiplied anything off to get their solutions. For instance, they may have started with factors like 4x – 4 or 5x – 15; I can't tell. So the general form of the family of quadratics (that is, the formula for every possible quadratic that has these zeroes) has to include any possible divided-off numbers. For this, I will use a letter to represent the divided-off numbers. The factors I know about are x – 1 and x – 3. Multiplying them together gives me x² – 4x + 3. Since some number may have been divided out, I have to multiply it back in:

The general form is a(x² – 4x + 3)

If they didn't divide anything out, then a will equal 1. By the way, your text may use some other letter for the constant; as long as you use some letter near the beginning of the alphabet (it's traditional), you should be fine.

• Find the quadratic with zeroes at x = 1 and x = 3, and
  passing through the point (0, –6).

Aha! With that extra point, I can narrow down the exact formula for the quadratic. The third point lets me account for that multiplier "a". I already know (from the previous problem) that the general form of any quadratic with these zeroes is a(x² – 4x + 3). But now I can plug in that point, and solve for a. Since (x, y) = (0, –6), then:

a((0)² – 4(0) + 3) = –6
3a = –6
a = –2   Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

Then "the" quadratic is –2(x² – 4x + 3) = –2x² + 8x – 6

• Find the quadratic with zeroes at x = ³/₂ and x = ^–5/₄, and
  passing through the point (½, 14).

Since the zeroes are x = ³/₂ and x = ^–5/₄, then the factors were x – ³/₂ and x – ( ^–5/₄ ) = x + ⁵/₄. Or... maybe the factors were 3x – 2 and 4x + 5... I really can't tell (not just from the zeroes, at least) whether they divided anything out when they were solving. So I'll need to use the point they gave me:

a(x – ³/₂ )(x + ⁵/₄ ) = y
a((½) – ³/₂ )((½) + ⁵/₄ ) = (14)
a(–1)( ⁷/₄ ) = 14
–( ⁷/₄ )a = 14
a = ( ¹⁴/₁ )( ^–4/₇ ) = –8

Then the exact quadratic is y = –8(x – ³/₂ )(x + ⁵/₄ )

If your book or teacher doesn't like fractions, you can multiply through, like this:

y = –8(x – ³/₂ )(x + ⁵/₄ )
y = –1(2)(x – ³/₂ )(4)(x + ⁵/₄ )
y = –1(2x – 3)(4x + 5)
y = –1(8x² – 2x – 15)
y = –8x² + 2x + 15

What if they ask for a quadratic, but only give you one root?

• Find the quadratic with a zero at x = sqrt(7) and passing through (2, –9).

When do I ever get square roots in my solutions to quadratics? When the stuff inside the square root in the Quadratic Formula doesn't simplify to a perfect square that I can take out. And the Quadratic Formula always spits out those messy square-root answers in opposite-signed pairs (from the "±" in front of the square root). So if x = sqrt(7) is a root, then so must x = –sqrt(7). And this gives me my two factors: x – sqrt(7) and x – (–sqrt(7)) = x + sqrt(7). Multiplying, I get x² – 7. I tack on my "I don't know if you divided anything out" constant a to get a(x² – 7). Plugging in the point they gave me, I get:

a(x² – 7) = y
a((2)² – 7) = (–9)
a(4 – 7) = –9
–3a = –9
a = 3

So the quadratic is y = 3(x² – 7) = 3x² – 21

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