Question

obtain all the zeroes of polynomial 2x4-7x3-13x2+63x-45 if two of its zeroes are 1 and 3​

Answer 1

Answer:

here is your solution ✌️

Answer 2

Answer:

-3 and $\frac{5}{2}$ are the other zeroes of the given polynomial

Step-by-step explanation:

Explanation:

Given , $2x^4- 7x^3- 13x^2 + 63 x- 45$ and two of its zeroes are given 1 and 3

So (x - 1 ) and (x - 3) will be the two factors of the given polynomial

Therefore ,

 (x-1)(x -3)  = $x^2 -4x + 3$

Step 1:

Now dividing $2x^4- 7x^3- 13x^2 + 63 x- 45$ by $x^2 -4x + 3$

$x^2 -4x + 3$ )$2x^4- 7x^3- 13x^2 + 63 x- 45$ (  $2x^2$ $+x - 15$

                  $2x^4 - 8x^3 + 6x^2$

                  -    +          -                        

                          $x^3 - 19 x^2 +63x -45$

                         $x^3 - 4x^2 + 3x$

                        -     +         -                        

                                -$15x^2 + 60 x-45$

                              - $15x^2 + 60 x-45$    

                               +       -          +                  

                               x         x         x

Step 2:

Therefore ,  other zeroes are ,

$2x^2$ $+x - 15$

⇒ $2x^2 + 6x -5x -15$     [By middle term splitting method ]

⇒2x(x+ 3)-5(x+3)

⇒(x+ 3)(2x -5)

∴x+ 3= 0 and 2x - 5 = 0

⇒ x = -3 and x = $\frac{5}{2}$

Final answer:

Hence , others zeroes are -3 and $\frac{5}{2}$ .

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