Question

If two zeros of the polynomial x⁴+7x³+-35x-60 are (-3) and (-4) and then find its third zero??



If someone tell the correct answer, I will surely marked as BRAINLIEST ANSWERS.​

Answer 1

Given zeros are nothing but (x+3) and (x+4).

So let p(x) = x⁴+7x³-35x-60

The product of the zeros will be (x+3)(x+4) = $x^{2} +7x +12$

Since p(x) will now be of the form Y( $x^{2} +7x +12$ ) where Y will be,

Y = [x⁴+7x³-35x-60] / ( $x^{2} +7x +12$ )

So, following the polynomial division, we get Y = $x^{2} - 5$

This can be done as follows,

                        x² - 5                

( $x^{2} +7x +12$ ) | x⁴+7x³-35x-60

                 ( - )  | $x^{4}$+7$x^{3}$+12x²            ------> Multiply with x² then subtract

                        |          -12x²-35x-60

                 ( - )  |           -12x²-35x-60  ------> Multiply with - 5 then subtract

                                            0

So our next zeros will be $x^{2} - 5$ = (x-5)(x+5)

So all roots of p(x) are (x-5)(x+5)(x+3)(x+4)

So the zeros of p(x) are 5, -5, -3, -4