Given the equation :
[tex]y=x^4+7x^3+25x^2-11x-150[/tex]to find the roots of he function , y = 0
so,
[tex]x^4+7x^3+25x^2-11x-150=0[/tex]the factors of 150 are;
1 x 150 , 2 x 75 , 3 x 50 , 5 x 30 ,
We will check which number give y = 0
so, when x = 1 , y = -128
When x = -1 , y = -120
when x = 2 , y = 0
So, x = 2 is one of the roots
so ( x - 2 ) is one of the factors of the given equation :
Make a long division to find the other roots:
so,
[tex]\frac{x^4+7x^3+25x^2-11x-150}{x-2}=x^3+9x^2+43x+75[/tex]See the following image:
Now , we will repeat the steps for the result
the factors of 75
1 x 75 , 3 x 25 , 5 x 5
We will check which number give y = 0
when x = 1 , y = 128
when x = -1 , y = 40
When x = 3 , y = 312
when x = -3 , y = 0
so, x = -3 is another root
So, ( x + 3 ) is one of the factors
so, make a long division again to find the other roots:
[tex]\frac{x^3+9x^2+43x+75}{x+3}=x^2+6x+25[/tex]See the following image :
Now the last function :
[tex]x^2+6x+25=0[/tex]a = 1 , b = 6 , c = 25
[tex]D=\sqrt[]{b^2-4\cdot a\cdot c}=\sqrt[]{36-4\cdot1\cdot25}=\sqrt[]{36-100}=\sqrt[]{-64}=i\sqrt[]{64}=\pm8i[/tex]which mean the last equation has no real roots
So,
the roots of the given equation is just two roots
So, the answer is the roots of the given eaution is x = 2 and x = -3
