Question

If two zeroes of polynomial P(x) = 2x4+ 10x3 + 5x2 - 35x - 42 are 3
and -2. Find other zeroes.
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Answer 1

The other two zeroes of the given polynomial

$P(x) = 2x^4+10x^3+5x^2-35x-42$

are     $x = (\sqrt{\frac{7}{2} }, -\sqrt{\frac{7}{2}})$  

• The given Polynomial is

$P(x) = 2x^4+10x^3+5x^2-35x-42$

• The given two zeroes of the above polynomial are (-2,-3)

         therefore we have x = -2 and x = -3 are two roots of the given       polynomial  

         This also means that (x + 2) and (x + 3) are two factors of the given polynomial

•  Also the product of these two factors is also a factor of P(x)

           that is,

            Q(x) =  $(x+2)(x+3)=x^2 +5x +6$ is a factor of P(x)

            Now, taking $x^2 +5x+6$ common from P(x), we get

                 $2x^4+10x^3+5x^2-35x-42 = 0$

                 $(x^2 +5x+6)(2x^2-7) = 0$

                      $2x^2-7 =0\\x^2 = \frac{7}{2}\\ x = \sqrt{\frac{7}{2} } , -\sqrt{\frac{7}{2} }$

• therefore the other two roots of the given polynomial P(x) are

                  $x = (\sqrt{\frac{7}{2} }, -\sqrt{\frac{7}{2}})$