How do you find all the zeros of f(x)=x^3+7x^2-6x-72 f(x)=x3+7x2−6x−72?
1 Answer
Use the rational root theorem to help find the first zero, then complete the square to find the other two, giving:
33 ,-4−4 ,-6−6
Explanation:
f(x) = x^3+7x^2-6x-72f(x)=x3+7x2−6x−72
By the rational root theorem, any rational zeros of
That means that the only possible rational zeros are:
+-1±1 ,+-2±2 ,+-3±3 ,+-4±4 ,+-6±6 ,+-12±12 ,+-18±18 ,+-24±24 ,+-36±36 ,+-72±72
Trying each in turn, the first one that works is:
f(3) = 3^3+(7*3^2)-(6*3)-72 = 27+63-18-72 = 0f(3)=33+(7⋅32)−(6⋅3)−72=27+63−18−72=0
So
x^3+7x^2-6x-72 = (x-3)(x^2+10x+24)x3+7x2−6x−72=(x−3)(x2+10x+24)
One way of factoring the remaining quadratic factor
x^2+10x+24x2+10x+24
=(x+5)^2-25+24=(x+5)2−25+24
=(x+5)^2-1^2=(x+5)2−12
=((x+5)-1)((x+5)+1)=((x+5)−1)((x+5)+1)
=(x+4)(x+6)=(x+4)(x+6)
Hence zeros:
x = -4x=−4 andx=-6x=−6
