Question

If f(x) = x3 + 12x2 +41x + 30 and f(-6) = 0, then find all of the zeros of f(x) algebraically.

Answer 1

$\text{Given that,}\\\\f(x) = x^3 +12x^2 +41x +30~~ \text{and} ~~ f(-6)=0\\\\\text{Since f(x)=0 when x = -6,}~ ~x-6 ~\text{is a factor of f(x).}\\\\x^3+12x^2+41x+30 = 0\\\\\implies x^3+6x^2+6x^2+36x+5x+30=0\\\\\implies x^2(x+6) +6x (x+6) +5(x+6)=0\\\\\implies (x+6)(x^2+6x+5)=0\\\\\implies (x+6)(x^2+5x+x+5)=0\\\\\implies (x+6)\textbf{[}x(x+5)+(x+5)\textbf{]}=0\\\\\implies (x+6)(x+1)(x+5)=0\\\\\text{Hence the roots are,}~ x = -6,~ x=-1~ \text{and}~ x = -5$

Answer 2

Answer:

-6, -5, -1

Step-by-step explanation:

Based on the question, (x+6) is a factor of f(x).

With this, you can do the long division method of dividing f(x) with (x+6) {Refer to the image}.

You'll get x²+6x+5 as a result.

Factorise it and you'll get

x²+6x+5 = (x+1)(x+5)

Then let (x+1)(x+5) = 0, x = -1, -5

Zero's of f(x) are -6, -5 -1