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Revision: Algebra >> Polynomials Maths English Medium Class 10 CBSE

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Definitions [2]

Definition: Geometrical Meaning of Zeroes

The zeroes of a polynomial are the x-coordinates of the points where the graph of y = p(x) intersects the x-axis.

Define a polynomial with real coefficients.

In the polynomial `f(x)=a_nx^n+a_(n-1)x^(n-1)+...+a_1x+a_0`, 

`a_nx^n,a_(n-1)x^(n-1),...,a_1x`, and `a_0` are known as the terms of the polynomial and `a_n,a_(n-1),...,a_1` and `a_0` are their real coefficients.

For example, `p(x)=3x-2` is a polynomial and 3 is a real coefficient.

Formulae [1]

Formula: Relation Between Zeroes (Roots) and Coefficients
Polynomial  Roots  Sum of roots  Pairwise product  Product of roots 
\[ax^2 + bx + c = 0\]
\[\alpha, \beta\]
\[-\frac{b}{a}\]
Not applicable
\[\frac{c}{a}\]
\[ax^3 + bx^2 + cx + d = 0\]
\[\alpha, \beta, \gamma\]
\[-\frac{b}{a}\]
\[\frac{c}{a}\]
\[-\frac{d}{a}\]
 

Theorems and Laws [2]

If α and β are the zeros of the quadratic polynomial f(x) = x2 – px + q, prove that `alpha^2/beta^2 + beta^2/alpha^2 = p^4/q^2 - (4p^2)/q + 2`.

Since α and β are the zeros of the quadratic polynomial f(x) = x2 – px + q

`alpha+beta=-"coefficient of x"/("coefficient of "x^2)`

`=(-(-p))/1`

= p

`alphabeta="constant term"/"coefficient of "x^2`

`=q/1`

= q

We have,

`alpha^2/beta^2+beta^2/alpha^2=(alpha^2xxalpha^2)/(beta^2xxalpha^2)+(beta^2xxbeta^2)/(alpha^2xxbeta^2)`

`alpha^2/beta^2+beta^2/alpha^2=alpha^4/(beta^2alpha^2)+beta^4/(alpha^2beta^2)`

`alpha^2/beta^2+beta^2/alpha^2=(alpha^4+beta^4)/(alpha^2beta^2)`

`alpha^2/beta^2+beta^2/alpha^2=((alpha^2+beta^2)-2alpha^2beta^2)/(alpha^2beta^2)`

`alpha^2/beta^2+beta^2/alpha^2=([(alpha+beta)^2-2alphabeta]^2-2(alphabeta)^2)/(alphabeta)^2`

`alpha^2/beta^2+beta^2/alpha^2=([(p)^2-2q]^2-2(q)^2)/q^2`

`alpha^2/beta^2+beta^2/alpha^2=([p^2-2q]^2-2q^2)/q^2`

`alpha^2/beta^2+beta^2/alpha^2=([p^2xxp^2-2xxp^2xx2q+2qxx2q]-2q^2)/q^2`

`alpha^2/beta^2+beta^2/alpha^2=([p^4-4p^2q+4q^2]-2q^2)/q^2`

`alpha^2/beta^2+beta^2/alpha^2=(p^4-4p^2q+4q^2-2q^2)/q^2`

`alpha^2/beta^2+beta^2/alpha^2=(p^4-4p^2q+2q^2)/q^2`

`alpha^2/beta^2+beta^2/alpha^2=p^4/q^2-(4p^2q)/q^2+(2q^2)/q^2`

`alpha^2/beta^2+beta^2/alpha^2=p^4/q^2-(4p^2)/q+2`

Hence, it is proved that `alpha^2/beta^2+beta^2/alpha^2" is equal to "p^4/q^2-(4p^2)/q+2`.

If the zeros of the polynomial f(x) = ax3 + 3bx2 + 3cx + d are in A.P., prove that 2b3 – 3abc + a2d = 0.

Let a - d, a and a + d be the zeros of the polynomial f(x). Then,

Sum of the zeroes `=("coefficient of "x^2)/("coefficient of "x^3)`

`a-d+a+a+d=(-3b)/a`

`3a=(-3b)/a`

`a=(-3b)/axx1/3`

`a=(-b)/a`

Since a is a zero of the polynomial f(x).

Therefore,

f(x) = ax3 + 3bx2 + 3cx + d

f(a) = 0

f(a) = aa3 + 3ba2 + 3ca + d

aa3 + 3ba2 + 3ca + d = 0

`a((-b)/a)^3+3bxx((-b)/a)^2+3cxx((-b)/a)+d=0`

`axx(-b)/axx(-b)/axx(-b)/a+3xxbxx(-b)/axx(-b)/a+3xxcxx(-b)/a+d=0`

`(-b^3)/a^2+(3b^3)/a^2-(3cb)/a+d=0`

`(-b^3+3b^3-3abc+a^2d)/a^2=0`

2b3 − 3abc + a2d = 0xa2

2b3 − 3abc + a2d = 0

Hence, it is proved that 2b3 − 3abc + a2d = 0.

Key Points

Key Points: Geometrical Meaning of the Zeroes of a Polynomial

A quadratic polynomial can have only:

Case (i): Two zeroes

  • Graph cuts the x-axis at two distinct points

  • Two distinct real zeroes

Case (ii): One zero

  • Graph touches the x-axis at exactly one point

  • One real (repeated) zero

Case (iii): No zero

  • The graph does not touch the x-axis at all

  • No real zero

A polynomial of degree n can have at most n zeroes.

Key Points: Relation Between Zeroes (Roots) and Coefficients of a Quadratic Equation
  • For \[ ax^{2} + bx + c = 0: \]
    \[ \alpha + \beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a} \]
  • A quadratic with roots \[ \alpha, \beta \] is:
    \[ x^{2} - (\alpha + \beta)x + \alpha\beta = 0 \]
  • For \[ ax^{3} + bx^{2} + cx + d = 0: \]
    \[ \alpha + \beta + \gamma = -\frac{b}{a} \]
    \[ \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} \]
    \[ \alpha\beta\gamma = -\frac{d}{a} \]
  • Always identify coefficients only after writing the equation in standard form.

Important Questions [29]

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