fxx4x328x276x48 factor the polynomial fx then solve the equa

f(x)=x^4-x^3-28x^2+76x-48 factor the polynomial f(x). then solve the equation f(x)=0

Solution

f(x)=x^4-x^3-28x^2+76x-48

find the zeros of the polynomial using rational root theorem :

he Rational Root Theorem tells you that if the polynomial has a rational zero then it must be a fraction pq, where p is a factor of the trailing constant and q is a factor of the leading coefficient.

The factor of the leading coefficient (1) is 1 .The factors of the constant term (-48) are 1 2 3 4 6 8 12 16 24 48 . Then the Rational Roots Tests yields the following possible solutions:

±1/1, ±2/1, ±3/1, ±4/1, ±6/1, ±8/1, ±12/1, ±16/1, ±24/1, ±48/1

Substitute the possibleroots one by one into the polynomial to find the actual roots. Start first with the whole numbers.

If we plug these values into the polynomial P(x), we obtain P(1)=0.

So, f(x)/(x-1) = (x^4-x^3-28x^2+76x-48)/(x-1)

= x^3 -28x +48

Solve  x^3 -28x +48 using rational root theorem :

he factor of the leading coefficient (1) is 1 .The factors of the constant term (48) are 1 2 3 4 6 8 12 16 24 48 . Then the Rational Roots Tests yields the following possible solutions:

±1/1, ±2/1, ±3/1, ±4/1, ±6/1, ±8/1, ±12/1, ±16/1, ±24/1, ±48/1

Substitute the possibleroots one by one into the polynomial to find the actual roots. Start first with the whole numbers.

If we plug these values into the polynomial P(x), we obtain P(2)=0.

Now find (  x^3 -28x +48)/(x-2) = x^2 +2x -24

solve the quadratic to get the roots which are x = 4 , -6

Roots of f(x) : x = 1 , 2 , 4 , -6

f(x) = (x-1)(x -2)(x-4)(x +6)