Question

1. Verify that 5, 1 / 2 , 3 / 4 are zeroes of cubic polynomial 8x^3 - 50x^2 + 53x - 15.

Class 10

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Answer 1

Given f(x) = 8x^3 - 50x^2 + 53x - 15.

Now,

f(5) = 8(5)^3 - 50(5)^2 + 53(5) - 15

      = 1000 - 1250 + 265 - 15

      = -985 + 985

      = 0.



f(1/2) = 8(1/2)^3 - 50(1/2)^2 + 53(1/2) - 15

         = 8(1/8) - 50(1/4) + 53/2 - 15

          = 1 - (25/2) + (53/2) - 15

         = -28 + 53 - 25

         = 0.



f(3/4) = 8(3/4)^3 - 50(3/4)^2 + 53(3/4) - 15

          = (27/8) - (225/8) + (159/4) - 15

          = (-99/4) + (159/4) - 15

          = -4 * 15 - 99 + 159

         = -60 - 99 + 159

         = -159 + 159

         = 0


Therefore 5, 1/2, 3/4 are the zeroes of the polynomial.


Verification:

On comparing with ax^3 + bx^2 + cx + d = 0, we get a = 8, b = -50, c = 53, d = -15.

Let p = 5, q = 1/2, r = 3/4.

= > p + q + r = 5 + 1/2 + 3/4

                     = 25/4.



Sum of zeroes = -b/a

= > p + q + r  = 50/8

(5) + (1/2) + (3/4) = 25/4

25/4  = 25/4


Hence, proved!


= > Sum of the product of zeroes = c/a

pq + qr + rp = c/a

(5)(1/2) + (1/2)(3/4) + (3/4)(5) = 53/8

(53/8) = 53/8


Hence proved!


= > Product of zeroes = -d/a

pqr = -d/a

(5)(1/2)(3/4) = -15/8

-15/8 = -15/8.


Hence proved!




Hope this helps!

Answer 2

Hey user

Here is your answer :-

$let \: p \:(x) = {8x}^{3} - 50 {x}^{2} + 53x - 15 \\ \\ p(5) = 8 {(5)}^{3} - 50 {(5)}^{2} + 53(5) - 15 \\ \\ = 125 \times 5 - 50 \times 25 + 53 \times 5 - 15 \\ \\ = 625 - 1250 + 265 - 15 \\ \\ = 875 - 1250 \\ \\ = - 385 \\ \\ if \: 5 \: is \: the \: factor \: p \: (5) \: should \: \: be \: 0 \: but \: p(x) \: not \: equal \: to \: 0 \\ \\ hence \: 5 \: is \: not \: a \: root \: of \: p \: (x)$

$if \: \frac{1}{2} \: is \:a \: root \: of \: p(x) then \: p( \frac{1}{2} ) \: should \: be \: equal \: to \: 0. \\ \\ p( \frac{1}{2} ) = 8 { x}^{3} - 50 {x}^{2} + 53 {x} - 15 \\ \\ = 8 { (\frac{1}{2} )}^{3} - 50 { (\frac{1}{2}) }^{2} - 53 (\frac{1}{2} ) - 15 \\ \\ = 1 - \frac{25}{2} + \frac{53}{2} - \frac{30}{2} \\ \\ = \frac{ - 4}{2} \\ \\ = - 2 \\ \\ so \: p( \frac{1}{2} ) \: is \: not \: eqaul \: to \: 0.$

$if \: \frac{3}{4} is \: root \: of \: p(x) \: then \: p( \frac{3}{4} )should \: be \: equal \: \: to \: 0 \\ \\ p( \frac{3}{4} ) = 8 {x}^{3} - 50 {x}^{2} + 53x - 15 \\ \\ = 8 {( \frac{3}{4} )}^{3} - 50 { (\frac{3}{4} )}^{2} + 53( \frac{3}{4} - 15 \\ \\ = \frac{27}{8} - \frac{225}{8} + \frac{159}{4} - 15 \\ \\ = \frac{ - 198}{8} + \frac{318}{8} - \frac{120}{8} \\ \\ =0 \\ \\ so \: p(x) \: is \: root \: of \: p(x) \:$

thank you.