Question

please answer this question ​

Answer 1

Answer:

the other two zeros are 7 and -5

Step-by-step explanation:

given 2 zeros are 2 + √3. and 2 - √3

x = 2 + √3 ;

x - 2 - √3 = 0

x = 2 - √3

x - 2 + √3 = 0

$(x - 2 - \sqrt{3} )(x - 2 + \sqrt{3} ) \\ = {(x - 2)}^{2} - {( \sqrt{3}) }^{2} \\ = {x }^{2} - 4x + 4 - 3 \\ = {x}^{2} - 4x + 1$

divide the given polynomial

$\frac{ {x}^{4} - 6 {x}^{3} - 26 {x}^{2} + 138x - 35 }{ {x}^{2} - 4x + 1} \\ = {x}^{2} - 2x - 35 \\ \\ {x}^{2} - 2x - 35 = 0 \\ {x}^{2} - 7x + 5x - 35 = 0 \\ x(x - 7) + 5(x - 7) = 0 \\ (x - 7)(x + 5) = 0 \\ x - 7 = 0 \: \: \: \: \: \: \: x + 5 = 0 \\ x = 7 \: \: \: \: \: \: \: \: \: \: \: \: \: \: x = - 5$

therefore the other two zeros are 7 and -5

hope you get your answer