Without using long division, or synthetic division, prove that expression x^2 + 5x + 6 is a factor of polynomial x^4 + 5x^3 + 2x^2 – 20x – 24.
x^2+5x+6 has factors of x+3 and x+2, so if their roots (x = -3, -2) are plugged into the polynomial in question, we should get 0 for both roots by the Remainder Theorem:
x^4 + 5x^3 + 2x^2 – 20x – 24
(-3)^4 + 5(-3)^3 + 2(-3)^2 - 20(-3) - 24
81 - 135 + 18 + 60 - 24
-54 + 18 + 60 - 24
-36 + 60 - 24
24 - 24
0 <-- x+3 is a factor of x^4 + 5x^3 + 2x^2 – 20x – 24
x^4 + 5x^3 + 2x^2 – 20x – 24
(-2)^4 + 5(-2)^3 + 2(-2)^2 - 20(-2) - 24
16 - 40 + 8 + 40 - 24
-24 + 8 + 40 - 24
-16 + 40 - 24
24 - 24
0 <-- x+2 is a factor of x^4 + 5x^3 + 2x^2 – 20x – 24
Thus, since both x+3 and x+2 are factors of x^4 + 5x^3 + 2x^2 – 20x – 24, then it is proved that x^2+5x+6 is a factor of x^4 + 5x^3 + 2x^2 – 20x – 24
The devisor factors to (x+2)(x+3)
So sub -2 and -3 into the quartic ,if you get 0 for both, you have proven it
To prove that the expression x^2 + 5x + 6 is a factor of the polynomial x^4 + 5x^3 + 2x^2 – 20x – 24, we can use the factor theorem.
According to the factor theorem, a polynomial P(x) has a factor of (x – a) if and only if P(a) = 0.
In this case, we need to show that P(x) = x^4 + 5x^3 + 2x^2 – 20x – 24 has a factor of (x^2 + 5x + 6).
Let us substitute x^2 + 5x + 6 into P(x):
P(x) = (x^2 + 5x + 6)^2 + 5(x^2 + 5x + 6) + 2(x^2 + 5x + 6) – 20x – 24
Expanding and simplifying:
P(x) = (x^4 + 10x^3 + 31x^2 + 60x + 36) + (5x^2 + 25x + 30) + (2x^2 + 10x + 12) – 20x – 24
P(x) = x^4 + 10x^3 + 38x^2 + 75x + 54
Since the coefficient of x^4 in P(x) is 1, and the coefficient of x^2 is 38, we can see that the expression x^2 + 5x + 6 is not a factor of P(x).
Therefore, it is proven that the expression x^2 + 5x + 6 is not a factor of the polynomial x^4 + 5x^3 + 2x^2 – 20x – 24.
To prove that the expression x^2 + 5x + 6 is a factor of the polynomial x^4 + 5x^3 + 2x^2 – 20x – 24 without using long division or synthetic division, we can make use of the Remainder Theorem.
The Remainder Theorem states that if a polynomial f(x) is divided by x - a, then the remainder is equal to f(a). In this case, instead of dividing by x^2 + 5x + 6, we can divide by x - (-2), since the expression x^2 + 5x + 6 factors into (x + 2)(x + 3), and -2 is one of the roots of the quadratic.
When we substitute x = -2 into the given polynomial x^4 + 5x^3 + 2x^2 – 20x – 24, we can check if the remainder is equal to zero. If it is, then the expression x^2 + 5x + 6 is indeed a factor of the polynomial.
By substituting x = -2 into the polynomial x^4 + 5x^3 + 2x^2 – 20x – 24, we get:
(-2)^4 + 5(-2)^3 + 2(-2)^2 – 20(-2) – 24
= 16 - 40 + 8 + 40 - 24
= 0
Since the remainder is equal to zero, we have proven that x^2 + 5x + 6 is a factor of x^4 + 5x^3 + 2x^2 – 20x – 24 without using long division or synthetic division.