### Chapter 2: Polynomials

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

(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 = ​ in equation (I), we get
2(​ )3 + (  )2 – 5( ​ ) + 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,  , 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 = ​ ​, β = 1,   = −2
Sum of the roots = α + β +  = ​ + 1 – 2 = ​ ​ – 1=  =
αβ + β  + α  =

Product of the roots = αβ   ​
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 = 2​ 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 = ​, β = 1,   = 1
Sum of the roots = α + β +  = 2 + 1 + 1 = ​4 =
αβ + β  + α  = 2​x1 + 1x1  + 2x1 ​  = 2 + 1 + 2 = 5
Product of the roots = αβ  = 1x1x1 ​  = 1 =
Therefore, the relationship between the zeroes and the coefficient are verified.

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