Finding Quadratic Zeroes and Verifying Coefficient Relationships

Hello! This first module is a balanced revision of core Class 10 mathematics ideas, beginning with polynomials. In this lesson, you will learn how to find the zeroes of a quadratic polynomial and then verify that those zeroes match the polynomial’s coefficients. This is a common board-exam format: find, show, and conclude clearly.

By the end, you should be able to work confidently with a quadratic polynomial in the form

$$p(x)=a{x}^2+bx+c,a\ne 0$$

and verify the two key relationships:

$$\alpha +\beta =-\frac{b}{a}$$ $$\alpha \beta =\frac{c}{a}$$

Here, $\alpha$ and $\beta$ are the zeroes.

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Zeroes: where a polynomial becomes zero

A zero of a polynomial $p(x)$ is a value of $x$ for which

$$p(x)=0$$

For example, consider

$$p(x)=x^2-4x-5$$

If $x=5$, then

$$p(5)=5^2-4(5)-5=25-20-5=0$$

So $5$ is a zero. Likewise, $x=-1$ is also a zero.

Graphically, zeroes are the $x$-coordinates where the graph meets the $x$-axis, because every point on the $x$-axis has $y=0$.

A quadratic polynomial can have at most two zeroes. On a graph, it may:

• cross the $x$-axis twice, giving two distinct zeroes;
• touch the $x$-axis once, giving two equal zeroes; or
• not meet the $x$-axis in the real plane.

For the questions in this lesson, factorisation will let us identify the zeroes directly.

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Finding zeroes by factorisation

To find the zeroes of a polynomial, first write the polynomial equal to zero and factorise it. Then use the zero-product property:

If a product of two factors is zero, at least one factor must be zero.

Take the polynomial

$$p(x)=2x^2-5x+3$$

We need to solve

$$2x^2-5x+3=0$$

Splitting the middle term

For a quadratic $a{x}^2+bx+c$, find two numbers whose:

• product is $ac$;
• sum is $b$.

Here,

$$a=2,b=-5,c=3$$

So we need two numbers with product

$$ac=2(3)=6$$

and sum $-5$. The numbers are $-2$ and $-3$.

Split the middle term:

$$2x^2-2x-3x+3=0$$

Group the terms:

$$2x(x-1)-3(x-1)=0$$

Factor out the common bracket:

$$(x-1)(2x-3)=0$$

Now set each factor equal to zero.

From

$$x-1=0$$

we get

$$x=1$$

From

$$2x-3=0$$

we get

$$x=\frac{3}{2}$$

Therefore, the zeroes are

$$\alpha =1,\beta =\frac{3}{2}$$

You can always check a result by substituting a zero back into the original polynomial. For example,

$$p(1)=2(1{)}^2-5(1)+3=0$$

So $1$ is confirmed as a zero.

A useful sign reminder

If a factor is $x+4$, then its zero is not $4$. Set it equal to zero:

$$x+4=0$$

so

$$x=-4$$

The sign changes when you solve the factor equation.

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Why zeroes are related to coefficients

Suppose a quadratic polynomial has zeroes $\alpha$ and $\beta$. Its factors must be

$$(x-\alpha )$$

and

$$(x-\beta )$$

Therefore, a quadratic with these zeroes can be written as

$$a(x-\alpha )(x-\beta )$$

where $a$ is the coefficient of $x^2$.

Expand the brackets:

$$a(x-\alpha )(x-\beta )$$ $$=a\left[x^2-(\alpha +\beta )x+\alpha \beta \right]$$ $$=a{x}^2-a(\alpha +\beta )x+a\alpha \beta$$

But the same polynomial is also written in standard form:

$$a{x}^2+bx+c$$

Compare the coefficients of matching terms:

Term	From the zeroes	From standard form
$x^2$ term	$a$	$a$
$x$ term	$-a(\alpha +\beta )$	$b$
Constant term	$a\alpha \beta$	$c$

So,

$$-a(\alpha +\beta )=b$$

Dividing by $-a$ gives

$$\alpha +\beta =-\frac{b}{a}$$

Also,

$$a\alpha \beta =c$$

Dividing by $a$ gives

$$\alpha \beta =\frac{c}{a}$$

These are not two unrelated formulas to memorise. They come from the fact that a quadratic can be expressed both through its coefficients and through its zeroes.

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The verification format to use in an exam

Return to the polynomial

$$2x^2-5x+3$$

We found its zeroes:

$$\alpha =1,\beta =\frac{3}{2}$$

Identify the coefficients carefully:

$$a=2,b=-5,c=3$$

1. Verify the sum of zeroes

Calculate the left-hand side using the zeroes:

$$\alpha +\beta =1+\frac{3}{2}$$ $$\alpha +\beta =\frac{5}{2}$$

Now calculate the right-hand side using coefficients:

$$-\frac{b}{a}=-\frac{-5}{2}$$ $$-\frac{b}{a}=\frac{5}{2}$$

Therefore,

$$\alpha +\beta =-\frac{b}{a}$$

2. Verify the product of zeroes

Calculate the left-hand side:

$$\alpha \beta =1\left(\frac{3}{2}\right)$$ $$\alpha \beta =\frac{3}{2}$$

Now use the coefficients:

$$\frac{c}{a}=\frac{3}{2}$$

Therefore,

$$\alpha \beta =\frac{c}{a}$$

Both relations have been verified.

A compact answer template

For any exam question of this type, organise your work like this:

1. Write $p(x)=0$.
2. Factorise.
3. State the zeroes clearly.
4. Identify $a$, $b$, and $c$, including their signs.
5. Calculate $\alpha +\beta$ and compare it with $-\frac{b}{a}$.
6. Calculate $\alpha \beta$ and compare it with $\frac{c}{a}$.
7. Write: Hence, the relationship between zeroes and coefficients is verified.

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Reading a graph and checking the coefficients

Look again at the graphed polynomial:

$$p(x)=x^2-4x-5$$

The graph shows the zeroes:

$$\alpha =-1,\beta =5$$

The coefficients are

$$a=1,b=-4,c=-5$$

Check the sum:

$$\alpha +\beta =-1+5=4$$ $$-\frac{b}{a}=-\frac{-4}{1}=4$$

Check the product:

$$\alpha \beta =(-1)(5)=-5$$ $$\frac{c}{a}=\frac{-5}{1}=-5$$

So the graph, the factorisation

$$x^2-4x-5=(x+1)(x-5)$$

and the coefficient relationships all describe the same polynomial.

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Common errors to avoid

Forgetting the negative sign in the sum formula.
The sum is

$$-\frac{b}{a}$$

not $\frac{b}{a}$. If $b=-4$, then $-b=4$.

Ignoring the leading coefficient.
For $2x^2-5x+3$, the product is not simply $3$. It is

$$\frac{c}{a}=\frac{3}{2}$$

Using the wrong numbers while splitting the middle term.
The two numbers must multiply to $ac$, not just $c$.

Not treating an absent term as zero.
For

$$x^2-9$$

the coefficient of $x$ is

$$b=0$$

Listing factors rather than zeroes.
If the factor is $2x-3$, write the zero as

$$x=\frac{3}{2}$$

not merely $2x-3$.

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You have now connected three views of the same idea: a zero as an input making $p(x)=0$, an $x$-intercept on a graph, and a value encoded in the coefficients of a quadratic polynomial.

The key facts to retain are:

$$\text{Sum of zeroes}=-\frac{b}{a}$$ $$\text{Product of zeroes}=\frac{c}{a}$$

and, when factorisation is possible, the zeroes come from setting each factor equal to zero. In the next lesson, you will build on this by deciding when factorisation is efficient and when the quadratic formula is the better way to solve a quadratic equation.