Quadratic Polynomial: Definition, Proof, Method & Graphs

Quadratic Polynomial: The quadratic formula is a formula that enables us to find the solutions of quadratic equations. A polynomial in \(x\) is an algebraic expression containing only non-negative integral powers of \(x\) when arranged in ascending or descending exponent of \(x\).

In other words, an algebraic expression of the form \({a_0} + {a_1}x + {a_2}{x^2} + \ldots + {a_{n – 1}}{x^{n – 1}} + {a_n}{x^n}\), where \({a_0},\,{a_1},\,{a_2} \ldots ,\,{a_n}\) are real numbers, \({a_n} \ne 0\) and \(n\) is a non-negative integer, is called a polynomial. \({a_0},\,{a_1},\,{a_2} \ldots ,\,{a_n}\) are called coefficients, and the highest power \(n\) of \(x\) is called index or power or radical or degree of the polynomial.

Thus, the highest power of \(x\) in a polynomial \(p(x)\) is called the degree of the polynomial \(p(x)\). A quadratic polynomial is of the form \(p(x) = a{x^2} + bx + c\), where \(a \ne 0\).

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Quadratic Polynomial: Definition

Polynomial is an expression of the form \(p(x) = {a_0} + {a_1}x + {a_2}{x^2} + \ldots + {a_n}{x^n}\), where \({a_n} \ne 0\), is called a polynomial in \(x\) of degree \(n\).

A polynomial is said to be linear, quadratic, cubic and biquadratic according to as its degree is \({\rm{1,}}\,{\rm{2,}}\,{\rm{3}}\) and \(4\) respectively. The general form of a quadratic polynomial is \(a{x^2} + bx + c\).

Value of a polynomial at a point: The value of a polynomial \(p(x)\) at \(x = \alpha \) is obtained by putting \(x = \alpha \) and it is denoted by \(p(\alpha )\).

Zeros of a polynomial: A real number \(\alpha \) is called a zero of \(p(x)\), if \(p(\alpha ) = 0\).

Quadratic Polynomial Notes

The zeros of a polynomial \(f(x)\) are all the \(x\) values that make \(f(x)\) equal to zero and a coefficient is an numeral that is multiplied with the unknown of a single term or the terms of a polynomial.

Relation between the zeros and coefficients of a quadratic polynomial:

Let \(\alpha \) and \(\beta \) be the zeros of a quadratic polynomial \(p(x) = a{x^2} + bx + c = 0\), where \(a \ne 0\).

Then, \((x – \alpha )\) and \((x – \beta )\) are the factors of \(p(x)\).

Therefore, \(\left( {a{x^2} + bx + c} \right) = k(x – \alpha )(x – \beta )\) Where \(k\) is a constant.

\( = k\left( {{x^2} – (\alpha + \beta )x + \alpha \beta } \right)\)

\( = k{x^2} – k(\alpha + \beta )x + k(\alpha \beta )\)

On comparing coefficients of like powers of \(x\) on both sides, we get \(k = a,\, – k(\alpha + \beta ) = b\) and \(k(\alpha \beta ) = c\).

\( \Rightarrow – a(\alpha + \beta ) = b\) and \(a(\alpha \beta ) = c[k = a]\)

\( \Rightarrow (\alpha + \beta ) = – \frac{b}{a}\) and \(\alpha \beta = \frac{c}{a}\)

\(\therefore {\rm{Sum}}\,{\rm{of}}\,{\rm{zeros}} = \frac{{ – \left( {{\rm{coefficient}}\,{\rm{of}}\,x} \right)}}{{\left( {{\rm{coefficient}}\,{\rm{of}}\,{x^2}} \right)}},\,{\rm{the}}\,{\rm{product}}\,{\rm{of}}\,{\rm{zeros}} = \frac{{{\rm{constant}}\,{\rm{term}}}}{{\left( {{\rm{coefficient}}\,{\rm{of}}\,{x^{\rm{2}}}} \right)}}\)

Quadratic Polynomial Method

There are various methods to factorise the quadratic polynomial, here we will discuss those methods in detail.

Factorisation of algebraic expression: The process of writing a given algebraic expression as the product of two or more factors is called factorisation of algebraic expressions.

Methods of factorisation of a quadratic polynomial:

1. Factorisation of a quadratic polynomial using standard identities
2. Factorisation of a quadratic polynomial utilising the method of splitting the middle term

1. Factorisation of a quadratic polynomial using standard identities

Some expressions can be factorised using the following identities.

\({(x + y)^2} = {x^2} + 2xy + {y^2}\)

\({(x – y)^2} = {x^2} – 2xy + {y^2}\)

\((x + y)(x – y) = {x^2} – {y^2}\)

In factorising, a given expression above given identities can be used whichever is required. First, we observe the given expression. If it is of \(RHS\) of any of the above-given identities, then \(LHS\) of that identity will be the required factors.

Problems based on factorisation of trinomials which are perfect squares:

Arrange the terms of the given trinomial in any of the following from whichever is suitable.

\({x^2} + 2xy + {y^2} = {(x + y)^2} = (x + y)(x + y)\)

\({x^2} – 2xy + {y^2} = {(x – y)^2} = (x – y)(x – y)\)

Problems based on factorisation of expressions which can be expressed in the form \(\left( {{x^2} – {y^2}} \right)\)

If the expression is of the form \(\left( {{x^2} – {y^2}} \right)\) use the formula \({x^2} – {y^2} = (x + y)(x – y)\)

If the expression is not of the form \(\left( {{x^2} – {y^2}} \right)\), try to transform the given expression in the form \(\left( {{x^2} – {y^2}} \right)\) and then use formula \({x^2} – {y^2} = (x + y)(x – y)\)

Go on using the formula \({x^2} – {y^2} = (x + y)(x – y)\) till the given expression is completely factored.

For example,

Factorise \({p^2} – 16\)

Answer: \({p^2} – 16 = {p^2} – {4^2}\)

\( = (p + 4)(p – 4)\)

Hence, the factorised form of the given expression is \((p + 4)(p – 4)\)

2. Factorisation of quadratic equation splitting the middle terms

Problems based on factorisation of the expression of the form \({x^2} + cx + d\)

If any factor is common in all terms of the given expression, take this factor out and write the remaining expression in brackets.

Find \(c\) which is the coefficient of \(x\) and \(d\) which is the constant term.

Find out factors \(a\) and \(b\) of \(d\) whose sum is \(c\).

That is, \(d = a \times b,\,c = a + b\)

Then, write \({x^2} + cx + d = {x^2} + (a + b)x + ab = \left( {{x^2} + ax} \right) + (bx + d)\) and take out common factor in each bracket.

How to find \(a\) and \(b\):

If the sign of \(q\) is positive then both the factors \(a\) and \(b\) of \(q\) will have the sign same as that of \(p\), i.e. if \(p\) is positive, then both \(a\) and \(b\) will be positive, but if \(p\) is negative, then both \(a\) and \(b\) will be negative.

If \(q\) is negative, then find \(|q|\) (Numerical value of \(q\)), then the numerically greater factor of \(q\) will have sign same as that of \(p\) and smaller factor will have sign opposite to that of \(p\).

Quadratic Polynomial Problems

In this section we will learn how to solve problems related to the value of the symmetric function in \(\alpha \) and \(\beta \), where \(a\)  and \(b\) are the zeroes of a quadratic polynomial, finding the zeros of a quadratic polynomial, finding a quadratic polynomial when its zeros are given.

Problems based on the value of the symmetric function in \(\alpha \) and \(\beta \), where \(a\)  and \(b\) are the zeroes of a quadratic polynomial.
Working Rule
1. First, find \(\alpha + \beta \) and \(\alpha \beta \).
For this, if \(a\) and \(b\) are zeroes of the quadratic polynomial \(a{x^2} + bx + c\), then \(\alpha + \beta = – \frac{b}{a}\) and \(\alpha \beta = \frac{c}{a}\)
2. Value of symmetric function in \(\alpha \) and \(\beta \) in terms of \(\alpha + \beta \) and \(\alpha \beta \).
For this use, the following relations depending on which is required.
(i) Express the given function in \(\alpha \) and \(\beta \) in terms of \(\alpha + \beta \) and \(\alpha \beta \). For this use, the following relations depending on which is required.
(ii) \({\alpha ^2} + {\beta ^2} = {(\alpha + \beta )^2} – 2\alpha \beta \)
(iii) \({(\alpha – \beta )^2} = {(\alpha + \beta )^2} – 4\alpha \beta \)
if \(\alpha \ge \beta \), then \(\alpha – \beta = \sqrt {{{(\alpha + \beta )}^2} – 4\alpha \beta } \)
\(|\alpha – \beta | = \sqrt {{{(\alpha + \beta )}^2} – 4\alpha \beta } \)
(iv) \({\alpha ^3} + {\beta ^3} = {(\alpha + \beta )^3} – 3\alpha \beta (\alpha + \beta )\)
(v) \({\alpha ^2} – {\beta ^2} = (\alpha + \beta )(\alpha – \beta ) = (\alpha + \beta )\sqrt {{{(\alpha + \beta )}^2} – 4\alpha \beta } ,\,\alpha \ge \beta \)

Problems based on finding the zeros of a quadratic polynomial.
Working Rule:
1. Write down the given quadratic polynomial in the form \(a{x^2} + bx + c\)
2. Find the factors of \(a{x^2} + bx + c\) by writing the middle term as the sum of two expressions. For this, if \(ac > 0\), then find two numbers \(p\) and \(q\) having same signs as that of \(b\) such that \(p + q = b\) and \(pq = ac\)
But, if \(ac > 0\), then find two numbers \(p\) and \(q\) having an opposite sign that \(p + q = b\) and \(pq = ac\).
3. Equate each factor thus obtained to zero to get the values of \(x\).
4. These values of \(x\) will be the zeros of the given quadratic polynomial.

Problems based on finding a quadratic polynomial when its zeros are given.
Working Rule:
1. A quadratic polynomial having \(\alpha \) and \(\beta \) as its zeroes is \({x^2} – ({\rm{sum}}\,{\rm{of}}\,{\rm{zeroes}})x + {\rm{product}}\,{\rm{of}}\,{\rm{zeroes}}\)
2. In fact, there are infinitely many quadratic polynomials having \(\alpha \) and \(\beta \) as their zeroes. They are: \(a\left[ {{x^2} – (\alpha + \beta )x + \alpha \beta } \right]\) where \(a\) is an arbitrary number.
We take a suitable number \(a\), usually the \(LCM\) of the denominators in the coefficients of \({x^2},\,x\) and constant term.

Quadratic Polynomial Graphical Representation

The zeroes of a quadratic polynomial \(a{x^2} + bx + c,\,a \ne 0\) are exactly the \(x\)-coordinates of the points where the parabola representing  \(y = a{x^2} + bx + c\) intersects the \(x\)-axis.

The parabola will open upwards if the leading coefficient is positive, and it will open downwards if it is negative.

Case (i): In this case, the graph cuts \(x\)-axis at two distinct points \(A\) and \(A’\). The \(x\)-coordinates of \(A\) and \(A’\) are the two zeroes of the quadratic polynomial \(a{x^2} + bx + c\) here.

Case (ii): In this case, the graph cuts the \(x\)-axis at exactly one point, i.e., at two coincident points.

The \(x\)-coordinate of \(A\) is the only zero for the quadratic polynomial \(a{x^2} + bx + c\) here.

Case (iii): In this case, the graph is either completely above the \(x\)-axis or completely below the \(x\)-axis. So, it will not cut the \(x\)-axis at any point.

So, the quadratic polynomial \(a{x^2} + bx + c\) has no zero here.

So, we can see geometrically that a quadratic polynomial can have either two distinct zeroes or two equal zeroes (i.e., one zero), or no zero. This also means that a polynomial of degree two has the utmost two zeroes.

Solved Examples – Quadratic Polynomial

Q.1. Find the zeros of the polynomial \(2{x^2} + 5x – 12\) and verify the relationship between its zeros and coefficients.
Ans: Coefficient of \({x^2} = 2\), coefficient of \(x = 5\) and the constant \( = – 12\)
Let the given polynomial be \(f(x)\). Then,
\(f(x) = 2{x^2} + 5x – 12\)
\( = 2{x^2} + 8x – 3x – 12\)
\( = 2x(x + 4) – 3(x + 4)\)
\( = (x + 4)(2x – 3)\)
\(\therefore \,f(x) = 0 \Rightarrow (x + 4)(2x – 3) = 0\)
\( \Rightarrow (x + 4) = 0\) or \((2x – 3) = 0\)
\( \Rightarrow x = – 4\) or \(x = \frac{3}{2}\)
So, the zeros of \(f(x)\) are \( – 4\) and \(\frac{3}{2}\).
Verification:
Sum of the zeros \( = \left( { – 4 + \frac{3}{2}} \right) = \frac{{ – 5}}{2} = \frac{{ – ({\rm{ coefficient}}\,{\rm{of }}\,x)}}{{\left( {{\rm{ coefficient}}\,{\rm{of }}\,{x^2}} \right)}}\)
and, the product of the zeros \( = ( – 4) \times \frac{3}{2} = \frac{{ – 12}}{2} = \frac{{{\rm{ constant}}\,{\rm{term }}}}{{\left( {{\rm{ coefficient}}\,{\rm{of }}\,{x^2}} \right)}}\)
Hence, the relationship between its zeros and coefficients are verified.

Q.2. Factorise \({y^2} + 10y + 25\).
Ans: The given algebraic expression is \({y^2} + 10y + 25\).
By suitably rewriting the terms of the given expression, we have:
\({y^2} + 10y + 25 = {y^2} + 2 \times 5 \times y + {5^2}\)
\( = {(y + 5)^2} = (y + 5)(y + 5)\)
Hence, the factorisation of \({y^2} + 10y + 25\) is \((y + 5)(y + 5)\).

Q.3. Factorise \({y^2} + 8y + 15\).
Ans:The given expression is \({y^2} + 8y + 15\)
Find two factors whose sum \( =8 \) and product \( =15 .\)
The factors are \(5\) and \(3\).
\(\therefore \,{y^2} + 8y + 15 = {y^2} + 5y + 3y + 15\)
\( = y(y + 5) + 3(y + 5)\)
\( = (y + 5)(y + 3)\)
Hence, the factorisation of \({y^2} + 8y + 15\) is \((y + 5)(y + 3).\)

Q.4. If \(\alpha \) and \(\beta \) are the zeroes of the quadratic polynomial \({x^2} + 5x + 4\), then find the values of (i) \(\frac{1}{\alpha } + \frac{1}{\beta }\), (ii) \({\alpha ^3} + {\beta ^3}\) (iii) \(\alpha – \beta \)
Ans: \(\alpha ,\,\beta \) are the zeroes of the quadratic polynomial \({x^2} + 5x + 4\)
\(\alpha + \beta = \frac{{ – {\rm{ (coefficient}}\,{\rm{of }}\,x)}}{{\left. {{\rm{ (coefficient}}\,{\rm{of }}\,{x^2}} \right)}} = – \frac{5}{1} = – 5,\,\alpha \beta = \frac{{{\rm{ constant}}\,{\rm{term }}}}{{\left. {{\rm{ (coefficient}}\,{\rm{of }}\,{x^2}} \right)}} = \frac{4}{1} = 4\)
(i) \(\frac{1}{\alpha } + \frac{1}{\beta } = \frac{{\alpha + \beta }}{{\alpha \beta }} = – \frac{5}{4}\)
(ii) \({\alpha ^3} + {\beta ^3} = {(\alpha + \beta )^3} – 3\alpha \beta (\alpha + \beta )\)
\( = {( – 5)^3} – 3 \times 4 \times ( – 5)\)
\( = – 125 + 60 = – 65\)
(iii) \(\alpha – \beta \)
\({(\alpha – \beta )^2} = {(\alpha + \beta )^2} – 4\alpha \beta \)
\( = {( – 5)^2} – 4 \times 4 = 9\)
Therefore, \(\alpha – \beta = \pm 3\).

Q.5. Find a quadratic polynomial with the given numbers as the sum and product of its zeros \(0,\,\sqrt 5 \)
Ans: Let \(\alpha ,\,\beta \) be the zeros of a quadratic polynomial.
Given, \(\alpha + \beta = 0\) and \(\alpha \beta = \sqrt 5 \)
Now, a quadratic polynomial whose zeros are \(\alpha \) and \(\beta \) is \({x^2} – (\alpha + \beta )x + \alpha \beta \)
\({x^2} – 0.x + \sqrt 5 \)
\( \Rightarrow {x^2} – \sqrt 5 \)
Hence, a quadratic polynomial is \({x^2} – \sqrt 5 \).

Summary

In this article, we learnt about the definition of a quadratic polynomial, quadratic polynomial notes, quadratic polynomial method, quadratic polynomial problems, quadratic polynomial graphical representation, solved examples on quadratic polynomial, frequently asked question on quadratic polynomial. The learning outcome of this article is, geometrically, that a quadratic polynomial can have either two distinct zeroes or two equal zeroes (i.e., one zero) or no zero. This also means that a polynomial of degree two has the utmost two zeroes.

Frequently Asked Question – Quadratic Polynomial

Let’s look at some of the commonly asked questions about quadratic polynomials:

Q.1. What is a quadratic polynomial? Give an example.
Ans: Quadratic polynomial is a polynomial of degree two is called a quadratic polynomial. \({x^2} + 2x + 3,\,2{x^2} – 3\) and \({y^2} – 13\) are few examples of a quadratic polynomial.

Q.2. How can you tell if a polynomial equation is quadratic?
Ans: Any equation of the form \(p(x) = 0\), where \(p(x)\) is a polynomial of degree \(2\), is a quadratic equation. But, when we write the terms of \(p(x)\) in the downward order of their degrees, then we obtain the standard form of the equation. That is, \(a{x^2} + bx + c = 0,\,a \ne 0\) referred to as the standard form of a quadratic equation.

Q.3. How do you write a quadratic polynomial?
Ans: The general form of a quadratic polynomial is \(a{x^2} + bx + c\), where \(a,\,b\) and \(c\) are real numbers and \(a \ne 0\).

Q.4. How many zeros can a quadratic polynomial have?
Ans: A quadratic polynomial can have either two distinct zeroes or two equal zeroes (i.e., one zero) or no zero. This also means that a polynomial of degree two has the utmost two zeroes.

Q.5. How do you write a quadratic polynomial when given zeros?
Ans: A quadratic polynomial having \(\alpha \) and \(\beta \) as its zeroes is \({x^2} – (\alpha + \beta )x + \alpha \beta \) i.e., \({x^2} – \left( {{\rm{sum}}\,{\rm{of}}\,{\rm{zeroes}}} \right)x + {\rm{product}}\,{\rm{of}}\,{\rm{zeroes}}{\rm{.}}\)

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