Question

Show that (x+4), (x-3) and (x-7) are factors of
x-6x²-19x+84.​

Answer 1

As f(-4) = 0, f(3) = 0 and f(7) = 0, this confirms that (x +4), (x - 3) and (x - 7) are factors of the polynomial x³ - 6x² - 19x + 84.

Explanation:

According to the Factor Theorem, if f(x) is a polynomial and f(a) = 0, then (x - a) is a factor of f(x). Therefore, to demonstrate that (x + 4), (x - 3), and (x - 7) are factors of x³ - 6x² - 19x + 84, substitute x = -4, x = 3, and x = 7 into the polynomial. If the result is zero for each substitution, it confirms that each of these expressions is indeed a factor of the polynomial.

`\begin{aligned}f(-4)&=(-4)^3-6(-4)^2-19(-4)+84\\&=-64-6(16)-19(-4)+84\\&=-64-96+76+84\\&=-160+76+84\\&=-84+84\\&=0\end{aligned}`

`\begin{aligned}f(3)&=(3)^3-6(3)^2-19(3)+84\\&=27-6(9)-19(3)+84\\&=27-54-57+84\\&=-27-57+84\\&=-84+84\\&=0\end{aligned}`

`\begin{aligned}f(7)&=(7)^3-6(7)^2-19(7)+84\\&=343-6(49)-19(7)+84\\&=343-294-133+84\\&=49-133+84\\&=-84+84\\&=0\end{aligned}`

Therefore, as f(-4) = 0, f(3) = 0 and f(7) = 0, this confirms that (x +4), (x - 3) and (x - 7) are factors of the polynomial x³ - 6x² - 19x + 84.