Verify that 1/2and 3 are zeroes of cubic polynomial 2x³-17x²+38x-15

2
by jyotiChander760

2015-09-10T12:37:03+05:30
Given cubic polynomial is 2x³-17x²+38x-15
Let x = 3
=2(3)³-17(3)²+38(3)-15
=54 - 153 + 114 - 15
=168 - 168
=0
Hence 3 is the zero of the given polynomial.
In the same way,
Let x = 1/2
=2(1/2)³-17(1/2)²+38(1/2)-15
=2(1/8) - 17(1/4) + 19 - 15
= 1/4 - 17/4 + 4
= -16/4 + 4
= -4 + 4
=0
Hence 1/2 is the zero of the given polynomial.
2015-09-10T18:50:42+05:30
If the zeroes of the polynomial p(x) 2x³ - 17x² + 38x - 15 are 1/2 and 3,
then x = 1/2 and 3.
When x = 1/2,
p(x) = 2*(1/2)³ - 17* (1/2)² + 38* 1/2 - 15
=> 2*1/8 - 17*1/4 + 38/2 -15
=>1/4 - 17/4 + 19 -15
=>1-17 + 4
4
=>-16 + 16
4
=>  0
4
=>  0
The value of the polynomial is  0. Therefore 1/2 is a zero of the polynomial.
If x = 3,
then p(x) = 2*(3)³ - 17* (3)² + 38*3 -15
⇒2*27 - 17* 9 +114 - 15
⇒54 - 153  + 114 - 15
⇒54 +114 - 153 - 15
⇒168 - 168
⇒0
The value is 0, therefore 3 is the zero of a polynomial.
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