Hi, please help with this it has to do with polynomial and real zeros.

Question

In Problems 1 and 2, find the real zeros, of f. Use the real zeros

1. f(x)=3x³+4x²+4x+1
2. f(x)=2x⁴+17x³+35x²-9x-45

Please do all the problems Thank You for your time :)

Answer 1

1)

Using the rational root theorem, possible roots are ±1/3, which is the last coefficient divided by the leading coefficient.

Trying f(-1/3) = 0 , so -1/3 is a zero. Dividing the polynomial by 3x+1 leaves x²+x+1 which is prime.

2) The trailing coefficient is -45 and the leading coefficient is 2. The factors of 2 are ±1,2. The factors of

45 are ±1,3,5,9,15,45, etc. Again using the rational root theorem, candidate factors are:

±45/1, 45/3,45/5, 45/9, 45/15, etc.

Trying -45/3=-5, f(-5) = 0, so -5 is a zero.

Dividing out x+5 leaves g(x) = 2x³+7x²-9.

Trying g(1) =0, so 1 is a zero. Dividing out x-1 leaves h(x)=2x²+9x+9, h(-3)=0, so -3 is another zero.

Dividing out x+3 leaves r(x) = 2x+3 of which x = -3/2 is a zero.

x=-5,1,-3,-3/2

For both problems 1 and 2, knowing synthetic division is extremely helpful.