We have to factorize f=x^3 + 4x^2 - 7x - 10

f=x^3 + 4x^2 - 7x - 10

=> x^3 + 6x^2 - 2x^2 + 5x - 12x - 10

=> x^3 + 6x^2 + 5x - 2x^2 - 12x - 10

=> x(x^2 + 6x +5) - 2(x^2 + 6x + 5)

=> (x - 2) ( x^2 + 6x +5)

=> (x - 2)(x^2 + 5x +x +5)

=> (x - 2)(x(x +5) +1( x + 5))

=> (x - 2)(x +5) (x +1)

Therefore x^3 + 4x^2 - 7x - 10 = **(x - 2)(x +5) (x +1)**

To factorise the polynomial f(x) = x^3+4x^2-7x-10.

We see that f(2) = 2^3+4*2^2-7*2-10 = 8+16-14-10 = 0.

Therefore we can write f(x) = (x-2)(x^2+kx+5) so that x^3 and constant terms are equal on both sides.

Put x = 1, then f(1) = 1^3+4*1^2-7*1-10 = (1-2)((1^2+k*1+5)

-12 = (-1)(6+k) = k-6. So k = 12-6 = 6.

Therefore x^3+4x^2-7x-10 =(x-2)(x^2+6x+5)....(1)

x^2-6x+5 = (x+5)(x+1).

So f(x) = x^3+4x^2-7x-10 = (x-2)(x+1)(x-5).

**Therefore the factors of the polynomial **

**x^3+4x^2-7x-10 = (x-2)(x+1)(x+5).**

First, we'll verify if x = -1 is the root of polynomial.

f(-1) = (-1)^3 + 4*(-1)^2 - 7*(-1) - 10

f(-1) = -1 + 4 + 7 - 10

We'll combine like terms and we'll have:

f(-1) = 0

If f(-1) = 0, then x = -1 is the root of polynomial.

We'll write the rule of division with reminder:

f(x) = (x + 1)(ax^2 + bx + c) + 0

The reminder is 0, since f(-1) = 0

x^3+4x^2-7x-10 = (x + 1)(ax^2 + bx + c)

We'll remove the brackets:

x^3+4x^2-7x-10 = ax^3 + bx^2 + cx + ax^2 + bx + c

Comparing, we'll get:

a = 1

a + b = 4

b = 4 - a

b = 3

b + c = -7

c = -7 - b

c = -10

ax^2 + bx + c = x^2 + 3x - 10

Now, we'l determine the roots of the quotient x^2 + 3x - 10.

x^2 + 3x - 10 = 0

x1 = [-3+sqrt(9 + 40)]/2

x1 = (-3+7)/2

x1 = 2

x2 = -5

The complete factorized polynomial is:

**f(x) = (x + 1)(x - 2)(x + 5)**