Question

one zero of the polynomial 3x^3+16x^2+15x-18 is 2/3.find the other zeros of the polynomial.(with explanation)​

Answer 1

Step-by-step explanation:

$\sf 3{x}^{3} + 16{x}^{2} + 15x - 18 = 0$

$\sf One\:of\:the\:zero\:is\: \frac {2}{3}$

$\sf so\:( 3x - 2 )\:should\:be\:the\:factor\:of\:the\:given\:polynomial$

$\sf So\:now\:try\:to\:create\:this\:factor$

$\sf 3 {x}^{3} - 2{x}^{2} + 18{x}^{2} - 12x + 27x - 18 = 0$

$\sf {x}^{2} ( 3x - 2 ) + 6x ( 3x - 2 ) + 9 ( 3x - 2 ) = 0$

$\sf ( 3x - 2 )( {x}^{2} + 6x + 9 ) = 0$

$\sf ( 3x - 2 )( x + 3 )( x + 3 ) = 0$

$\sf So\:the\:other\:zero\:of\:the\:polynomial\:is\:-3$

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