Question

verify -2/3,1/4,2/5 are the zero of polynomial 60x^3+x^2-20x+4​

Answer 1

Solution:-

(1).For P(-2/3),

=) p(x) = 60x³ + x² - 20x + 4

=) p( -2/3) = 60 (-2/3)³ + (-2/3)² - 20(-2/3) + 4

=) p = 60 × (-8)/27 + 4/9 + 40/3 + 4

=) p = -20×8/9 + 4/9 + 40/3 + 4

=) p = [ (-160) + 4 + 120 + 36]/9

=) p = -32.

Hence,

P(-2/3) is not the zero of the Given polynomial.

(2). For p(1/4).

=) p(x) = 60x³ + x² - 20x + 4

=) p(1/4) = 60. (1/4)³ + (1/4)² - 20/4 + 4

=) p(1/4) = 60/64 + 1/16 - 5 + 4

=) p(1/4) = 15/16 + 1/16 -1

=) p(1/4) = [ 15 + 1 - 16]/16

=) p(1/4) = (16-16)/16 = 0.

Hence,

p(1/4) is the zero of the Given polynomial.

(3). For p(2/5).

=) p(x) = 60x³ + x² - 20x + 4

=) p(2/5) = 60(2/5)³ + (2/5)² - 20(2/5) + 4

=) p(2/5) = 60×( 8/125) + 4/25 - 8 + 4

=) p(2/5) = 12 × 8/25 + 4/25 - 4

=) p(2/5) = [ 96 + 4 - 100]/25

=) p(2/5) = (100 - 100)/25 = 0.

Hence,

p(2/5) is the zero of the Given polynomial.