Question

find all the zeros of polynomial x 4 + x cube minus 14 x square - 2 X + 24 if two of its zeros are root 2 and minus root 2​

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

$f(x)=x^{4}+x^{3}-14x^{2}-2x+24$

First, given zeroes are $\sqrt{2}$ and $-\sqrt{2}$

Thus the factors are $(x-\sqrt{2})$ and $(x+\sqrt{2})$

($\therefore{g}(x)=(x-\sqrt{2})(x+\sqrt{2})\\\\\Rightarrow{g}(x)=x^{2}-{(\sqrt{2})}^{2}\\\\\Rightarrow{g}(x)=x^{2}-2$

$f(x)=x^{4}+x^{3}-14x^{2}-2x+24\\\\\Rightarrow{f}(x)=x^{4}-14x^{2}+24+x^{3}-2x\\\\\Rightarrow{f}(x)=x^{4}-2x^{2}-12x^{2}+24+x^{3}-2x\\\\\Rightarrow{f}(x)=x^{2}(x^{2}-2)-12(x^{2}-2)+x(x^{2}-2)\\\\\Rightarrow{f}(x)=(x^{2}-2)(x^{2}-12+x)\\\\\Rightarrow{f}(x)=(x^{2}-2)(x^{2}+x-12)$

Now

$q(x)=\frac{f(x)}{g(x)}=\frac{(x^{2}-2)(x^{2}+x-12)}{(x^{2}-2)}= (x^{2}+x-12)$

Now using middle term break we get,

$x^{2}+x-12=0\\\\\Rightarrow{x}^{2}+4x-3x-12=0\\\\\Rightarrow{x}(x+4)-3(x+4)=0\\\\\Rightarrow(x+4)(x-3)=0$

therefore the value of x is

$x+4=0\\\\\Rightarrow{x}=-4$

$(x-3)=0\\\\\Rightarrow{x}=3$

Thus, the four Zeroes are $\sqrt{2}$,$-\sqrt{2}$,-4 and 3