EVALUATING AND FINDING ZEROS OF POLYNOMIALS

Evaluating Polynomials

Question 1 :

Find the value of the polynomial 

f(y)  =  6y - 3y² + 3 at 

(i) y  =  1 (ii)  y  =  -1  (iii)  y  =  0

Solution :

f(y)  =  6y - 3y² + 3 at 

(i) y  =  1

f(1)  =  6(1) - 3(1)² + 3

  =  6 - 3 + 3

  =  6

(ii)  y  =  -1

f(-1)  =  6(-1) - 3(-1)² + 3

  =  -6 - 3 + 3

  =  -6

(iii)  y  =  0

f(0)  =  6(0) - 3(0)² + 3

  =  3

Question 2 :

If p(x) = x² – 2√2 x + 1, find p(2√2)

Solution :

p(x) = x² – 2√2 x + 1

p(2√2)  =  (2√2)² – 2√2 (2√2) + 1

  =  4(2) - 4(2) + 1

  =  8 - 8 + 1

  =  1

Finding Zeros of Polynomials

To find zeroes of a polynomial, we have to equate the  polynomial to zero and solve for the variable.

Question 2 :

Find the zeros of each polynomial.

(i) p(x) = x – 3

Solution :

p(x)  =  0

x - 3  =  0

x  =  3

(ii) p(x) = 2x + 5

Solution :

p(x)  =  0

2x + 5  =  0

2x  =  -5

x  =  -5/2

(iii) q(y) = 2y – 3

Solution :

q(y)  =  0

2y - 3  =  0

2x  =  3

x  =  3/2

(iv)  f(z)  =  8z 

Solution :

f(z)  =  0

8z  =  0

z  =  0

(iv) p(x)  =  ax, when a ≠ 0 

Solution :

p(x)  =  0

ax  =  0

x  =  0

(vi) h(x) = ax + b, a ≠ 0, a,b ∈ R

Solution :

h(x)  =  0

ax + b  =  0

ax  =  -b

x  =  -b/a

Question 3 :

Find the roots of the polynomial equations.

(i) 5x – 6 = 0

Solution :

5x - 6  =  0

5x  =  6

x  =  6/5

(ii) x + 3 = 0

Solution :

x + 3  =  0

x  =  -3

(iii) 10x + 9 = 0

Solution :

10x + 9  =  0

10x  =  -9

x  =  -9/10

(iv) 9x – 4 = 0

Solution :

9x - 4  =  0

9x  =  4

x  =  4/9

(v) 2x² - x + (1/5)

Solution :

(10x² − 5x + 1)/5  =  0

10x² − 5x + 1  =  0

10x² − 10x  + 5x + 1  =  0

10x (x + 1) + 1(5x + 1)  =  0

(10x + 1)   (5x + 1)  =  0

10x + 1  =  0     (or)  5x + 1  =  0

10x  =  -1    (or)  5x  =  -1

x  =  -1/10   (or)  x  =  -1/5

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