i cant figure these out, its been 3 hours now ?

1) P(x) = x^3+7x^2+10x+4

Look at all possible factors of 4, and see if any of them when substituted for x will give 0. All the terms are positive, so positive factors can be ignored (e.g. 1 + 7*1 + 10*1 + 4 = 22), but

P(-1) = -1 + 7 - 10 + 4 = 0

That means x + 1 is a factor (because x = -1 makes x+1 = zero)

Now if you've been taught "synthetic long division" of polynomials, use that to divide P(x) by x+1 to find the other (quadratic) factor. Otherwise

P(x) = (x+1)(x² + ...?)

Expanding this gives us x³ + x² + ...

and since we want 7x², to add to the x² we already have we put 6x in the second parentheses:

P(x) = (x+1)(x² + 6x +...?)

Expanding that gives x³ + 6x² + x² + 6x +

Now, since we want 10x, we need an extra 4x, so we put 4 in the parentheses:

P(x) = (x+1)(x² + 6x + 4), and if you check by expanding you'll find it gives the original polynomial.

Now the solutions of P(x) = 0 are

x + 1 = 0 or x² + 6x + 4 = 0

i.e. x = -1 or x = -3 +/- √5 using the quadratic formula.

2) x^4-7x^3+42x-36=0

This time, if there are integer zeros they will be some of 1, 2, 3, 4, 6, 9, 12, 18, 36 and their negatives. Luckily, x = 1 is a zero (it's the first value we try out):

1 - 7*1 + 42*1 - 36 = 0

Therefore x - 1 is a factor.

Now using the same process as before,

x^4-7x^3+42x-36 = (x - 1)(x³ - 6x² - 6x + 36)

Now try various values of x in x³ - 6x² - 6x + 36:

2³ - 6*2² - 6*2 + 36 = 8 so that doesn't work

6³ - 6*6² - 6*6 + 36 = 0 and so x - 6 is a factor.

x³ - 6x² - 6x + 36 = (x - 6)(x² - 6)

and x² - 6 = (x + √6)(x - √6)

Hence the equation becomes

(x - 1)(x - 6)(x + √6)(x - √6) = 0

Therefore

x = 1, 6, √6 or -√6

3) you mean x^4+6x^3+13x^2+24x+36 = 0 (otherwise it isn't an equation). Using the same processes as before,

x^4+6x^3+13x^2+24x+36

= (x + 3)(x³ + 3x² + 4x + 12)

= (x + 3)(x + 3)(x² + 4)

Do you use complex numbers? The solutions are

x = -3 (with multiplicity 2, because the factor (x+3) occurs twice) , -2i, 2i.

The only real solution is x = -3.

For the first problem note that all terms are positive, so any real zeros must be negative. By the rational roots theorem the only possible rational roots are now the negative divisors of 4, namely: -1, -2, or -4. Check P(-1) and find that P(-1) = 0, therefore (x+1) is a factor of P(x). Divide P(x) by (x+1), either manually or via synthetic division and get:

P(x) = (x+1)(x^2+6x+4). The remaining zeros are the roots of x^2+6x+4 which can be found by completing the square:

x^2 + 6x = -4

x^2 + 6x + 9 = -4 + 9 = 5

(x+3)^2 = 5, so

x+3 = sqrt(5) or x+3 = -sqrt(5), so finally

x = -3 + sqrt(5) or x= -3 - sqrt(5).

Summary: the zeros of P(x) are: -1, -3 + sqrt(5), -3 - sqrt(5).

That's it for now. Good luck on the other two problems.