Chapter 2: Polynomials

Q&A -Ask Doubts and Get Answers

Q
Polynomials CBSE NCERT Solutions

Question:

Verify that the numbers given along side of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:

  1.  2x3 + x2 - 5x + 2;    1/2, 1, -2
  2.  x3 - 4x2 + 5x – 2;   2, 1, 1

Answer:

(i) 2x3 + x2 - 5x + 2;  1/2, 1, -2

Suppose p(x) = 2x3 + x2 - 5x + 2   ……….. (I)

Now we verify , 1 and -2 are zeroes of the polynomial

On substituting the value of x = โ€‹ in equation (I), we get
2(frac{1}{2}โ€‹ )3 + ( โ€‹frac{1}{2} )2 – 5(frac{1}{2} โ€‹ ) + 2

 = frac{1}{4} + frac{1}{4} - frac{5}{2} + 2

= frac{1+1-10}{4} + 2

= frac{-8}{4} + 2

= -2 + 2 = 0
On substituting the value of x = 1 in equation (I), we get
2(1)3 + (1)2 – 5(1) + 2

= 2 + 1 – 5 + 2

= 5 - 5 = 0
On substituting the value of x = -2 in equation (I), we get
2(-2)3 + (-2)2 – 5(-2) + 2

= 2(-8) + 4 + 10 + 2

= -16 + 16 = 0

Therefore, frac{1}{2} , 1 and -2 are the zeroes of the given polynomial.
Now we can Compare the polynomial 2x3 + x2 - 5x + 2   with ax3 + bx2 + cx + d, we obtain,

= 2,b = 1,c = −5, d = 2
Let us assume α = โ€‹ โ€‹, β = 1,  \gamma = −2
Sum of the roots = α + β + \gamma = โ€‹frac{1}{2} + 1 – 2 = โ€‹ frac{1}{2}โ€‹ – 1= -\frac{1}{2} = -\frac{b}{a}
αβ + β\gamma  + \gammaα  = โ€‹  \frac{1}{2} \times 1 + 1\times(-2) +\frac{1}{2} \times (-2)

                        = \frac{1}{2} - 2 - 1 = \frac{1}{2} - 3 = \frac{-5}{2}= \frac{c}{a}
Product of the roots = αβ\gamma   โ€‹ = \frac{1}{2} \times 1 \times(-2) = -1 = - \frac{d}{a}
Therefore, the relationship between the zeroes and the coefficient are verified.

 

 

(ii) x3 - 4x2 + 5x – 2;   2, 1, 1

Suppose p(x) = x3 - 4x2 + 5x – 2  ……….. (I)

Now we verify 2, 1 and 1 are zeroes of the polynomial.

On substituting the value of x = 2โ€‹ in equation (I), we get
     23 – 4(2)2 + 5(2) – 2 

 8 – 16 + 10 – 2

= 18 – 18 = 0

On substituting the value of x = 1 in equation (I), we get
   13 – 4(1)2 + 5(1) – 2 

= 1 - 4 + 5 - 2

= 6 - 6 = 0
Therefore, 2, 1 and 1 are the zeroes of the given polynomial.
Now we can Compare the polynomial x3 - 4x2 + 5x – 2 with ax3 + bx2 + cx + d, we obtain,

= 1,b = -4, c = 5, d = -2
Let us assume α = โ€‹, β = 1,  \gamma = 1
Sum of the roots = α + β + \gamma = 2 + 1 + 1 = โ€‹4 = -\frac{b}{a}
αβ + β\gamma  + \gammaα  = 2โ€‹x1 + 1x1  + 2x1 โ€‹  = 2 + 1 + 2 = 5  \frac{c}{a}
Product of the roots = αβ\gamma  = 1x1x1 โ€‹  = 1 = -\frac{d}{a}
Therefore, the relationship between the zeroes and the coefficient are verified.

VIDEO EXPLANATION


Related Questions for Study

What our students and parents say about us!

Choose EduSakshamยฎ
Embrace Better Learning

edusaksham-mobile-app