Because there are no logical roots, the sole option is to factorize into two factors of degree 2. Because the polynomial is monic, the components are also monic. As a result, the probable factors are $x2 + ax + b$ and $x2 + cx + d$ with $a,b,c,dinmathbb Z$. $$ f = (x2 + ax + b)(x2 + cx + d) = x4 + (a + c)x3 + (ac + b + d)x2 + (ad + bc)x + bd $$ When we compare the coefficients, we obtain four equations. $$ a+c = 12 ac + b + d = 44 ad + bc = 72 bd = 36 ad + bc = 72 bd = 36 $$ The final equation has solutions $$ (b,d)in(1,36),(-1,-36),(2,18),(-2,-18),(3,12),(-3,-12),(4,9),(-4,-9),(6,6),(-6,-6) up to switching the two factors. $$ It takes some effort, but inserting these numbers into the remaining three equations reveals that there is only one solution for $b=d=6$, which is $a = 4$, $b = 8$, or $a = 8$, $b=4$. This gives you the two variables.

### 5(X-1)+2(X+3)+6=0 Verify

If (f(x))2 = 1, then f(x) = 1 or f(x) = -1. Fix x 0, then use your argument to determine if f(x 0) = 1 or f(x 0) = -1. If f(x 0)=1, we will show that f(x)=1 for every other x. If y 0 exists by contradiction such that f(y 0)=-1...

### 16. 5(X-1)+2(X+3)+6=0

Because there are no logical roots, the sole option is to factorize into two factors of degree 2. Because the polynomial is monic, the components are also monic. As a result, the probable factors are $x2 + ax + b$ and $x2 + cx + d$ with $a,b,c,dinmathbb Z$. $$ f = (x2 + ax + b)(x2 + cx + d) = x4 + (a + c)x3 + (ac + b + d)x2 + (ad + bc)x + bd $$ When we compare the coefficients, we obtain four equations. $$ a+c = 12 ac + b + d = 44 ad + bc = 72 bd = 36 ad + bc = 72 bd = 36 $$ The final equation has solutions $$ (b,d)in(1,36),(-1,-36),(2,18),(-2,-18),(3,12),(-3,-12),(4,9),(-4,-9),(6,6),(-6,-6) up to switching the two factors. $$ It takes some effort, but inserting these numbers into the remaining three equations reveals that there is only one solution for $b=d=6$, which is $a = 4$, $b = 8$, or $a = 8$, $b=4$. This gives you the two variables.

3,1 All rational roots of a polynomial have the form racpq, where p divides the constant term -3 and q divides the leading coefficient 1. List all potential candidates racpq. x=1 Find one such root by testing out all the integer values, beginning with the least absolute value and working your way up. If no integer roots are discovered, experiment with fractions.

### X^6+X^5+X^4+X^3+X^2+X+1=0 Mcq

x^{3}+2x^{2}-2x-1=0 According to the Factor theorem, x-k is a polynomial factor for each root k. Divide x-4+x-3-4x2+x+1 by x-1 to obtain x-3+2x2-2x-1. Solve the problem such that the outcome is 0.1. All rational roots of a polynomial have the form racpq, where p divides the constant term -1 and q divides the leading coefficient 1. List all potential candidates racpq.