$$2$$ and $$-2$$ are two zeros of the polynomial $$m^4+m^3-34m^{2} - 4m + 120$$. What are the other two zeros of the polynomial?
$${ m }^{ 4 }+{ m }^{ 3 }-34{ m }^{ 2 }-4m+120=f\left( m \right) $$
$$2$$ & $$-2$$ are two zeroes of polynomial
Let $$\alpha ,\beta $$ be other two zeroes
Sum of roots$$=\alpha +\beta +2-2=-1=\cfrac { -Coeff.\quad of\quad { m }^{ 3 } }{ Coeff.of{ m }^{ 4 } } $$
$$\alpha +\beta =-1\longrightarrow 1$$
Product of roots$$=\alpha \beta \left( 2 \right) \left( -2 \right) =\cfrac { constant\quad term }{ Coeff.of{ m }^{ 4 } } $$
$$-4\alpha \beta =120$$
$$\alpha \beta =-30\longrightarrow 2$$
$${ \left( \alpha -\beta \right) }^{ 2 }={ \left( \alpha +\beta \right) }^{ 2 }-4\alpha \beta $$
$${ \left( \alpha -\beta \right) }^{ 2 }={ \left( -1 \right) }^{ 2 }-4\left( -30 \right) $$
$${ \left( \alpha -\beta \right) }^{ 2 }=1+120$$
$$\alpha -\beta =\pm 11$$
$$I.\alpha -\beta =11\longrightarrow 3$$
Add 1 & 3
$$2\alpha =10$$
$$\alpha =5$$
$$\therefore \beta =-6$$
$$II.\alpha -\beta =-11\longrightarrow 4$$
Add 1 & 4
$$2\alpha =-12$$
$$\alpha =-6$$
$$\beta =5$$
$$\therefore 5$$ and $$-6$$ are other two zeroes.
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