Any Algebraic expression with constants and variables is known as a Polynomial. Polynomial is a combination of two words, “Poly” and “nominal” where poly means “many” and nominal means “terms”, hence, a Polynomial can contain as many terms but never infinite. In a Polynomial, zeros and coefficients are present where the coefficients are already provided to us however, we need to obtain the value of zeros with the help of coefficients.

### Relationship between Zeros and Coefficients

We know that zeros of any polynomial are the points where the graph of the polynomial cuts the x-axis. These zeros can also be found out using the coefficients of different terms in a polynomial. Let’s look at the relationship between the zeros and coefficients of the polynomials.

**Linear Polynomial**

A linear polynomial, in general, is defined by,

y = ax + b

We know, for zeros we need to find the points at which y = 0. Solving this general equation for y = 0.

y = ax + b

⇒0 = ax + b

⇒x =

This gives us the relationship between zero and the coefficient of a linear polynomial.

In general for a linear equation y = ax + b, a ≠ 0, the graph of ax + b is a straight line that cuts the x-axis at (, 0)

**Question: Verify the zeros of the linear polynomial both using the formula mentioned above and the graphical method. **

**y = 4x + 2**

**Solution:**

We are given with the equation y = 4x + 2,

Here a = 4 and b = 2

So, by the formula mentioned above the zero will occur at (, 0) that is

Let’s verify this zero with graphical method. We need to plot the graph of this equation.

y = 4x + 2

Let’s bring it to the intercept form.

Now we know the intercepts on the x and y-axis.

**Quadratic Polynomial**

Quadratic polynomials have the highest degree of 2 and along with factorization, there are other methods through which the zeros of the quadratic polynomial can be found such as **Dharacharya method**. As it has the highest degree 2, there exist 2 zeros in a quadratic polynomial.

Let’s derive the relationship between zeros and coefficients of a quadratic polynomial. Let’s assume a polynomial,

P(x) = 2x^{2} – 8x + 6

This polynomial can be factorized as follows,

P(x) = 2x^{2} -8x + 6

= 2x^{2} – 6x – 2x + 6

= 2x(x – 3) -2(x – 3)

= 2(x – 1)(x – 3)

So this equation has roots x = 1 and x = 3

Notice that,

Sum of zeros = 1 + 3 = 4 =

Product of zeros = 1 × 3 = 3 =

This is the relationship between zeros and coefficients for second-order coefficients.

So to put in a general form

For a polynomial, p(x) = ax

^{2}+ bx + c which has m and n as rootsm + n =

m × n =

**Question: Verify the property stated above for the equation, 6x ^{2} – 10x + 4**

**Solution:**

6x

^{2}– 10x + 4⇒ 6x

^{2}– 6x -4x + 4⇒ 6x(x – 1) – 4(x – 1)

⇒ (6x – 4) (x – 1)

Thus, the roots for this equation come out to be x = and x = 1

Now we know according to above properties,

Sum of zeros =

Product of zeros =

Let’s verify it

Sun of zeros = 1 + =

Both the values come to be equal. Hence, verified.

Let’s verify the product of roots property

Product of zeros =

In this case also, both values are equal. Hence, verified.

**Cubic Polynomial**

Similar relation can be derived for a cubic polynomial. A cubic polynomial is a polynomial of degree 3 and since it has its highest degree as 3, there exist three zeros of a cubic polynomial. Let’s suppose the roots/zeros of the polynomials obtained are p, q, r, the relationship between the zeros and polynomials will be given as,

For a cubic polynomial,

ax

^{3}+ bx^{2}+ cx + dWhich has roots x = p, q and r

p + q+ r =

pq + qr + pr =

pqr =

Let’s look at some examples regarding these properties.

### Sample Problems

**Question 1: Verify that -1,1,2 are the roots of the polynomial x ^{3} -2x^{2} – x + 2. Also**,

**verify the properties stated above.**

**Solution:**

x = -1, 1 and 2 are the roots of polynomial that is P(x) = 0 at all these points. Let’s plug in the values of x one by one in the polynomial.

P(-1) = (-1)

^{3}– 2(-1)^{2}-(-1) + 2= -1 – 2 + 1 + 2

= -3 + 3

= 0

P(1) = (1)

^{3}– 2(1)^{2}– 1 + 2= 1 – 2 – 1 + 2

= 0

P(2) = (2)

^{3}– 2(2)^{2}– 2 + 2= 8 – 8 – 2 + 2

= 0

Thus, all these points are roots of the polynomial

Let’s verify the properties

(1) p + q+ r =

⇒ -1 + 1+ 2 =

⇒ 2 = 2

L.H.S = R.H.S

Hence Verified

(2) pq + qr + pr =

⇒ (-1)(1) + (1)(2) + (-1)(2) =

⇒ -1+ 2 -2 = -1

⇒ -1 = -1

L.H.S = R.H.S

Hence Verified

(3) pqr =

⇒ (-1)(1)(2) =

⇒ -2 = -2

L.H.S = R.H.S

Hence Verified

**Question 2: Evaluate the sum and product of the zeros of quadratic polynomial 6x ^{2} + 18. **

**Solution: **

General form of a quadratic polynomial is ax

^{2}+ bx + c = 0.Given polynomial 6x

^{2}+ 18 can be rewritten as,6x

^{2}+ 0.x + 18As studied above the sum and product of roots of quadratic polynomial is given by,

m + n =

m.n =

Where m and n are the roots of the polynomial

In our case a = 6, b = 0 and c = 18. Plugging the values in the formulas

m + n =

m.n =

Thus, the sum of roots is 0 and product is given by 3.

**Question 3: Given a polynomial ax ^{2} + bx + 1. Its roots are -1 and 3. Find the values of a and b. **

**Solution:**

We know the formula for sum and product of the root of a quadratic polynomial .

Let m and n be the roots,

Here, m = -1 and n =3

So, by the formula

m + n = …(1)

m.n = …..(2)

From the equation (1)

-1 + 3 =

⇒ 2 = \frac{-b}{a}

⇒ 2a = -b

From equation (2)

(-1)(3) =

⇒ -3a = 1

⇒ a =

Putting this value of “a” in the above equation

b =