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If two zeros of tje polynomial x4_6x3_26x2+138x_38 are 2 +_root3 find other zeroes

Posted by Ankit Pandey 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

Two zeros are {tex}2\pm\sqrt3{/tex}
Sum of Zeroes {tex}2 + \sqrt { 3 } + 2 - \sqrt { 3 } = 4{/tex}
and product of zeroes = {tex}( 2 + \sqrt { 3 } ) ( 2 - \sqrt { 3 } ) = 4 - 3 = 1{/tex}
Hence quadratic polynomial formed out of this will be a factor of given polynomial,
So, x² - (sum of zeroes)x + product of zeroes
= x² - 4x + 1 will be a factor of given polynomial,
Divide given polynomial with x² - 4x + 1 to get other zeroes.

Now,
x² -2x - 35
= x² - 7x + 5x - 35
= x(x - 7) + 5(x - 7)
= (x - 5) (x - 7)
{tex}\therefore {/tex} Zeros are
x = 7 and x = -5
{tex}\therefore {/tex} Other two zeros are 7 and -5

0Thank You

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CBSE > Class 10 > Mathematics