Welcome! Today we're tackling Scaling to Integer Coefficients and the Verification Loop — a crucial skill for cleaning up your polynomial answers.

Many construction problems produce polynomials with fractional coefficients.

The standard presentation expects integer coefficients.

You need a systematic way to clear fractions without changing the zeros.

What changes	What stays the same
Coefficients	Zeros
Appearance	Roots of the equation

In this lesson, you will master:

The LCM scaling procedure — the systematic technique
The construction-verification loop — catching errors before they go unnoticed

1. The LCM scaling procedure

Scaling: The Clean-Up Step 🧹

Scaling is the clean-up step after construction. The procedure is simple:

Find the LCM of all denominators
Multiply every term by it

But the 'every term' part is critical — including terms that already have integer coefficients.

📋 Given Info

Given Information:

For $x^{2} - \frac{7}{6} x - \frac{1}{2}$ , the denominators are 1, 6, and 2.

$LCM (1, 6, 2) = 6$

Multiply every term by 6LCM.

✍️ Question

Your Turn ✏️

Scale $x^{2} - \frac{7}{6} x - \frac{1}{2}$ to have integer coefficients.

Show your work: multiply each term by the LCM (which is 6) and write the final polynomial.

The scaling procedureSystematically clearing fractions:

List all denominators: $x^{2}$ has denominator 11Even the hidden ones like 1, $\frac{7}{6} x$ has denominator 6, $\frac{1}{2}$ has denominator 2.

Find the LCMThis single number eliminates all fractions: $LCM (1, 6, 2) =$ $6$ key!Will eliminate all fractions at once

Multiply EVERY term by 6Not just the terms with fractions:
- $6 \times x^{2} =$ $6 x^{2}$
- $6 \times (- \frac{7}{6} x) = - \frac{7 \times 6}{6} x =$ $- 7 x$
- $6 \times (- \frac{1}{2}) = - \frac{6}{2} =$ $- 3$

✍️ MCQ

Choose one

When we multiplied

6 \times \left(-\frac{7}{6}x\right)

, why did the fraction disappear?

✓

Result: $6 x^{2} - 7 x - 3$

We've successfully converted our fractional quadratic into one with integer coefficientsNo fractions left, factorisation becomes straightforward!

✍️ T/F

True or False?

The scaled polynomial

6x^2 - 7x - 3

has the same roots as the original

x^2 - \frac{7}{6}x - \frac{1}{2}

. True or False?

✓

⚠️ Common mistake: Multiplying only the fractional terms by 6, leaving $x^2$ This is the key mistake to avoid unchanged.

This gives $x^2 - 7x - 3$ wrong!The roots won't match anymore, which is a completely different polynomial.

✍️ Yes/No

Yes or No?

If we multiply only the fractional terms by 6 (leaving

x^2

unchanged), we get

x^2 - 7x - 3

. Will this polynomial have the same roots as the original

x^2 - \frac{7}{6}x - \frac{1}{2}

?

✓

When you scale, EVERY termIncluding ones with integer coefficients gets multiplied — including the ones that already have integer coefficients.

Term	Correct scaling	Wrong approach
$x^{2}$	$6 \times x^{2} = 6 x^{2}$	Left as $x^{2}$ ❌
$- \frac{7}{6} x$	$6 \times (- \frac{7}{6} x) = - 7 x$	$- 7 x$ ✓
$- \frac{1}{2}$	$6 \times (- \frac{1}{2}) = - 3$	$- 3$ ✓

So always ask yourself — did I multiply EVERY term by the same number?If any term was skipped, the polynomial changes

2. Why scaling preserves zeros

🤔 A Natural Question

When working with polynomials, we often need to multiply them by constants to clear fractions and get integer coefficients.

But here's something worth thinking about:

If you multiply a polynomial by 6, do the zeros change?

Understanding why they do (or don't) is important for mathematical confidence — and for knowing when scaling is safe.

📋 Given Info

Here's a key fact to keep in mind:

If $p (α) = 0$ , then for any constant $k$ (where $k \neq 0$ ):

k \cdot p(\alpha) = k \cdot 0 = 0

Scaling a polynomial by a non-zero constant preserves all its zeros.

✍️ Question

Your Turn 📝

A student worries:

"If I multiply the polynomial by 6, won't it have different zeros?"

Explain why this concern is unfounded.

Here is why scaling is safeMultiplying by a constant preserves zeros.

Let $p (x) = x^{2} - \frac{7}{6} x - \frac{1}{2}$ . Suppose $\alpha$ is a zeroAlpha makes the polynomial equal zero of this polynomial, meaning $p(\alpha) = 0$ This is the condition we're preserving.

Now create a new polynomial by scalingScaling to get integer coefficients: $q (x) = 6 \cdot p (x) = 6 x^{2} - 7 x - 3$ .

What happens when we plug in $α$ ?

$q (α) = 6 \cdot p (α) = 6 \cdot 0 = 0$

See? If $p (α) = 0$ , then $6 \cdot p (α)$ is just $6 \times 0$ , which is still zerokey!Zero multiplied by anything is still zero!

The zero stays a zero!We don't lose or gain any roots

✍️ MCQ

Choose one

If

p(\alpha) = 0

and we define

q(x) = 10 \cdot p(x)

, what is

q(\alpha)

?

✓

Let's evaluate $q$ at $\alpha$ :

q(\alpha) = 6 \cdot p(\alpha) = 6 \cdot 0 = 0

So $α$ is also a zero of $q (x)$ . The zeros are identical.

This is the key principleThis is the essential principle to keep in mind: multiplying a polynomial by any non-zero constant $k$ preserves all its zerosScaling preserves all the roots.

✍️ MCQ

Choose one

If

p(\alpha) = 0

and we define

q(x) = 100 \cdot p(x)

, what is

q(\alpha)

?

✓

This works because multiplying zero by any number gives zeroZero times anything is still zero. As long as the constant is not 0Zero multiplier destroys everything (which would make the polynomial vanish entirely), scaling preserves all zerosAll roots remain unchanged.

✍️ MCQ

Choose one

If

p(\alpha) = 0

, what is

k \cdot p(\alpha)

for any non-zero constant

k

?

✓

Note: Scaling by a NEGATIVE constantSign doesn't matter for zeros also works. If you multiply by $-6$ still worksAny non-zero value works, the zeros are still the same. Only multiplying by $0$ would destroy information.

3. The construction-verification loop

The Construction-Verification Loop 🔄

The final skill is the verification loop: after constructing a polynomial, factor it back and check that you recover the original zeros.

This circular check catches sign errorswatch out! and arithmetic mistakescareful! that would otherwise go unnoticed.

Circular flow diagram showing 5 steps in a loop: 1. Compute sum and product -> 2. Construct polynomial -> 3. Scale to integers -> 4. Factor -> 5. Verify zeros -> back to step 1, with arrows connecting each step in a cycle

The Complete Loop:

Compute sum and product from given zeros
Construct using the formula $p(x) = x^2 - (\text{sum})x + (\text{product})$
Scale to integer coefficients
Factor the scaled polynomial
Verify the zeros match the original

This circular process is your built-in error-checking system.

✍️ Question

Your Turn ✏️

Construct a quadratic polynomial whose zeros are $\frac{5}{2}$ and $1$ .

Complete the full verification loop:

Scale to integer coefficients
Then factor the result to verify you get back the original zeroscheck

Let's do the full loop.

Step 1: Zeros are $\frac{5}{2}$ and $1$ The verification loop: find zeros, sum, product, check coefficients.

Finding SumConnects directly to the coefficient of x (S):

$S = \frac{5}{2} + 1 = \frac{5}{2} + \frac{2}{2} =$ $\frac{7}{2}$ SThis links to the x coefficient in your polynomial

Finding ProductConnects to the constant term (P):

$P = (\frac{5}{2}) (1) =$ $\frac{5}{2}$ PThe product relates to the constant term

denominator 2Both have denominator 2, so multiply by 2 to clear fractions

✍️ MCQ

Choose one

Using the formula

p(x) = x^2 - (\text{sum})x + (\text{product})

, what is the coefficient of

x

in the unscaled polynomial?

✓

Step 2: Construct.

$p(x) = x^2 - \frac{7}{2}x + \frac{5}{2}$ The formula pattern you use every single time

Step 3: Scale. Denominators: 2 and 2. LCM = 2keyFind the LCM, multiply through, and you're done.

$2x^2 - 7x + 5$ Scale to get integer coefficients, zeros stay the same

✍️ Yes/No

Yes or No?

If we had scaled by 4 instead of 2, would the zeros of the polynomial change?

✓

Step 4: Factor the polynomial $2 x^{2} - 7 x + 5$

We use the splitting the middle termYour go-to technique when coefficient of x squared is not 1 method.

First, find $a \cdot c = 2 \times 5 = 10$

We need two numbers that:

Multiply to give $10$ First condition - numbers must multiply to give a times c
Add to give $-7$ Second condition - must add to give the middle coefficient (the coefficient of $x$ )

The pair is: $- 5$ and $- 2$ ✓

✍️ MCQ

Choose one

Why do we need the two numbers to multiply to

10

specifically (and not some other number)?

✓

Factoring by grouping:

$2 x^{2} - 7 x + 5$

$= 2 x^{2} - 5 x - 2 x + 5$ (splitting $- 7 x$ as $- 5 x - 2 x$ )

$= x (2 x - 5) - 1 (2 x - 5)$ (grouping)If they don't match, something went wrong in splitting

(factored) $= (2x - 5)(x - 1)$ The matching bracket lets you factor it out ✓

✍️ MCQ

Choose one

From the factored form

(2x - 5)(x - 1) = 0

, what are the two zeros?

✓

Step 5: Zeros. $2 x - 5 = 0$ gives $x = \frac{5}{2}$ . $x - 1 = 0$ gives $x = 1$ .

These match the original zeros. Construction verified. ✓Matching zeros confirms your construction is correct

✍️ MCQ

Choose one

Why is the verification step (checking that the zeros match) necessary in the construction-verification loop?

✓

If the zeros had NOT matched, there would be an error somewhere — most likely a sign error in the construction formula or a mistake in computing $S$ or $P$ .

Common culprits:

Forgetting the negative signSigns are where most students slip up in $x^2 - Sx + P$ The minus before S is the common mistake
Arithmetic errors when adding fractions for $S$
Sign errors when multiplying for $P$ (Check signs in sum and product calculations)