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Solved Examples - Polynomials, Class 10, Mathematics Notes | Study Additional Documents & Tests for Class 10 - Class 10

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Ex.1 Find the zeros of the quadratic polynomial x2 + 7x + 12, and verify the relation between the zeros and its coefficients.
 Sol. 
We have,
f(x) = x2 + 7x + 12 = x2 + 4x + 3x + 12
Solved Examples - Polynomials, Class 10, Mathematics Notes | Study Additional Documents & Tests for Class 10 - Class 10 f(x) = x(x + 4) + 3 (x + 4)
Solved Examples - Polynomials, Class 10, Mathematics Notes | Study Additional Documents & Tests for Class 10 - Class 10 f(x) = (x + 4) (x + 3)
The zeros of f(x) are given by f(x) = 0
Solved Examples - Polynomials, Class 10, Mathematics Notes | Study Additional Documents & Tests for Class 10 - Class 10 x2 + 7x + 12 = 0
Solved Examples - Polynomials, Class 10, Mathematics Notes | Study Additional Documents & Tests for Class 10 - Class 10 (x + 4) (x + 3) = 0
Solved Examples - Polynomials, Class 10, Mathematics Notes | Study Additional Documents & Tests for Class 10 - Class 10 x + 4 = 0 or, x + 3 = 0
Solved Examples - Polynomials, Class 10, Mathematics Notes | Study Additional Documents & Tests for Class 10 - Class 10 x = – 4 or x = – 3
Thus, the zeros of f(x) = x2 + 7x + 12 are α = – 4 and β = – 3
Now, sum of the zeros =α+β  = (–4) + (–3) = – 7

Solved Examples - Polynomials, Class 10, Mathematics Notes | Study Additional Documents & Tests for Class 10 - Class 10

Ex.2 Find the zeros of the quadratic polynomial f(x) = abx2   + (b2 + ac) x + bc and verify the relationship between the zeros and its coefficients.
 Sol.
f(x) = abx2   + (b2 + ac) x + bc = abx2   + b2x + acx + bc
= bx (ax + b) + c (ax + b) = (ax + b) (bx + c)
So, the value of f(x) is zero when ax + b = 0 or bx + c = 0, i.e. x = –b/a or x = –c/b
Therefore, –b/a and –c/b are the zeros (or roots) of f(x).
Solved Examples - Polynomials, Class 10, Mathematics Notes | Study Additional Documents & Tests for Class 10 - Class 10
Solved Examples - Polynomials, Class 10, Mathematics Notes | Study Additional Documents & Tests for Class 10 - Class 10
Solved Examples - Polynomials, Class 10, Mathematics Notes | Study Additional Documents & Tests for Class 10 - Class 10
Solved Examples - Polynomials, Class 10, Mathematics Notes | Study Additional Documents & Tests for Class 10 - Class 10

Ex.3 Find a quadratic polynomial each with the given numbers as the sum and product of its zeros respectively.
 (i) 14 , –1 (ii) 2 , 13
 Sol. 
We know that a quadratic polynomial when the sum and product of its zeros are given is given by –
f(x) = k {x2   – (Sum of the zeros) x + Product of the zeros}, where k is a constant.
(i) Required quadratic polynomial f(x) is given by f(x) = k (x2   – 1/4 x – 1 )

(ii) Required quadratic polynomial f(x) is given by f(x) = k (x2   –  x 1/3 )

Ex.4 Divide the polynomial 2x2   + 3x + 1 by the polynomial x + 2 and verify the division algorithm.
 Sol.
We have

Solved Examples - Polynomials, Class 10, Mathematics Notes | Study Additional Documents & Tests for Class 10 - Class 10

Clearly, quotient = 2x – 1 and remainder = 3
Also, (x + 2) (2x – 1) + 3 = 2x2   + 4x – x – 2 + 3 = 2x2  + 3x + 1
i.e., 2x2   + 3x + 1 = (x + 2) (2x–1) + 3. Thus, Dividend = Divisor × Quotient + Remainder.

Ex.5 Check whether the polynomial t2 – 3 is a factor of the polynomial 2t4 + 3t3 – 2t2 – 9t – 12, by dividing the second polynomial by the first polynomial.
 Sol.
We have

Solved Examples - Polynomials, Class 10, Mathematics Notes | Study Additional Documents & Tests for Class 10 - Class 10

Since the remainder is zero, therefore, the polynomial t2 – 3 is a factor of the polynomial 2t4 + 3t3 – 2t2 – 9t – 12

Ex.6 Find all the zeros of 2x4 – 3x3 – 3x2 + 6x – 2, if you know that two of its zeros are √2 and – √2.
 Sol. 
Let p(x) = 2x4 – 3x3 – 3x2 + 6x – 2 be the given polynomial. Since two zeros are √2 and – √2 so, (x –√2 ) and (x+√2 ) are both factors of the given polynomial p(x).
Also, (x – √2) (x + √2) = (x2 – 2) is a factor of the polynomial. Now, we divide the given polynomial by x2 – 2.

Solved Examples - Polynomials, Class 10, Mathematics Notes | Study Additional Documents & Tests for Class 10 - Class 10

Solved Examples - Polynomials, Class 10, Mathematics Notes | Study Additional Documents & Tests for Class 10 - Class 10  

Hence, all the zeros of the polynomial 2x4 – 3x3 – 3x2 + 6x – 2 are √2 , – √2 , 1 and 1/2

Ex.7 On dividing f(x) = x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and – 2x + 4,
 respectively. Find g(x).
 Sol.
Here, Dividend = x3 – 3x2 + x + 2
Quotient = x – 2,
Remainder = – 2x + 4 and Divisor = g(x).
Since Dividend = Divisor × Quotient + Remainder
So, x3 – 3x2 + x + 2 = g(x) × (x –2) + (–2x + 4)
Solved Examples - Polynomials, Class 10, Mathematics Notes | Study Additional Documents & Tests for Class 10 - Class 10 g(x) × (x – 2) = x3 – 3x2 + x + 2 + 2x – 4

Solved Examples - Polynomials, Class 10, Mathematics Notes | Study Additional Documents & Tests for Class 10 - Class 10

Ex.8. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following :
 (i) p(x) = x3 – 3x
  + 5x – 3, g(x) = x2 – 2
 (ii) p(x) = x4 – 3x
 + 4x + 5, g(x) = x2 + 1 – x
 (iii) p(x) = x4 – 5x + 6, g(x) = 2 – x
2  .
 

Sol.

Solved Examples - Polynomials, Class 10, Mathematics Notes | Study Additional Documents & Tests for Class 10 - Class 10

Hence, Quotient q(x) = x – 3 and Remainder r(x) = 7x – 9

Ex.9. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the
 first polynomial.
 (i) t
– 3, 2t4 + 3t– 2t2 – 9t – 12 (ii) x+ 3x + 1, 3x4 + 5x– 7x+ 2x + 2
 (iii) x3 – 3x + 1, x5 – 4x
+ x+ 3x + 1

Sol:

Solved Examples - Polynomials, Class 10, Mathematics Notes | Study Additional Documents & Tests for Class 10 - Class 10

Hence, t2– 3 is a factor of 2t4+ 3t3 – 2t2 – 9t – 12

Ex.10. Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
 (i) deg p(x) = deg q(x)
 (ii) deg q(x) = deg r(x)
 (iii) deg r(x) = 0.

Sol. (i) p(x) = 2x2 + 2x + 8, g(x) = 2x0 = 2; q(x) = x2 + x + 4 ; r(x) = 0
(ii) p(x) = 2x2 + 2x + 8 ; g(x) = x2 + x + 9 ; q(x) = 2 ; r(x) = – 10
(iii) p(x) = x3 + x + 5 ; g(x) = x2 + 1 ; q(x) = x ; r(x) = 5.

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