What is the number of turning points in the graph of the function of x defined below?

y = 2x² + 5x - 7

Solution:

The polynomial y has degree n=2 so the maximum number of turns it can have is n-1 = 2-1 = 1. (In fact we know that our polynomial is a quadratic and it opens upwards so this is another way to know it only has one turn).

The polynomial equation x(x2+4)(x2-x-6)=0 has how many real roots?

Solution:

x2+4 contributes complex zeros. This means that the factors x and x2-x-6 contribute real zeros. Therefore the polynomial has 3 real zeros

Find all zeros of the polynomial: P(x)=(x^2+4)(x^2+x)-(x^2+4)

 

solutions are
, 
and

Find the exact zeros (show your work using synthetic division and\ or factoring )F(x)=x^3-7x^2+13x-3.

What is the leading coefficient of -3x^2+2x+5x^3 ?

The leading coefficient is the coefficient of the term of greatest degree (regardless of the position of that term) so it is 5

How do you use the intermediate value theorem to show the polynomial function that has a zero in the given. [1.3, 1.9], f(x)= x^5-x^4+3x^3-4x^2-16x +15

f(1.3)=-5.112 (approx.)
f(1.9)=2.465 (approx.)
the theorem tells us that since there is a sign change for f on the interval there must be a zero on the interval

find the number a such that x + 2 is a factor of 4x^3 + ax^2 -5x - 2

synthetic division leads us to the terms
4
a-8
11-2a
4a-24
so 4a-24=0 leads to a=6

What is the solution for x? sqrt(x+5)=x^2-5

The solution is very long a little tough...but I will summarize...

square both sides to get:

Simplify and rewrite as

We wish to write this as a product of square polynomials, so after some careful thought and experimentation we use



When we expand this version and compare coefficients we can determine that

1) 
2) 
3) 

This nonlinear can be solved by substitution. It has a rational solution of




therefore we can write:


and that means


so that


and we know there are only two solutions so...

---

Find the polynomial of least degree having zeros: -5i and 6+i

Assuming you are familiar with the theory... 

-5i is a root of x^2+25

6+i comes from x^2-12x+37

...their product is... 

(x^2+25)(x^2-12x+37)=
x^4-12x^3+62x^2-300x+925

(and any constant multiple of this will due)

If you are given the root, 2-3i, how would you write that as a factor? If this is confusing then, for example, a root is 1 so then the factor would be (x-1), so what would it be for a complex number?

As a factor we'd write:
x-(2-3i)
i.e.
x-2+3i

Use the given zero to find all the zeros of the function: f(x)=x^3+x^2+9x+9 zero: 3i i=imaginary number

since 3i is a root, then so is -3i and thus so is x^2+9. Then factoring we can say...

so...

{-1, 3i, -3i}

Solve the equation 3x^3+x^2-8x+4=0

by synthetic division we can determine that

then by some other method we can see that


therefore the solution is:
{ 1, 2/3, -2}

subtract (6a^3+a-3)-(4-3a^3-6a^2)

(6a^3+a-3)-(4-3a^3-6a^2)=
6a^3+a-3-4+3a^3+6a^2=
9a^3+6a^2+a-7

Use the Rational Zero Test to list all possible rational zeros of f. f(x)=x^3-4x^2-4x+16

since the leading coefficient is 1, we only look at the constant terms factors
16:
1, -1, 2, -2, 4, -4, 8, -8, 16, -16
in fact toe actual zeros are -2, 2, and 4

Find the complex zeros of each polynomial function. write f in factored form. f(x)=x^4+ 2x^3 + 22x^2 +50x- 75

x^4+ 2x^3 + 22x^2 +50x- 75=
(2x^3 +50x)+(x^4 + 22x^2 - 75)=
2x(x^2 +25)+(x^4 + 25x^2 -3x^2- 75)=
2x(x^2 +25)+(x^4 + 25x^2) -(3x^2+ 75)=
2x(x^2 +25)+x^2(x^2 + 25) -3(x^2+ 25)=
(x^2 +25)(x^2+2x-3)=
(x^2+25)(x+3)(x-1) 

zeros: {-5i, 5i, -3, 1} 



:)

when combining like terms what is the answer to (x^3+2)^2

After expanding...

True or False: A Polynomial can be described as a finite linear combination of power functions.

True. 
Power functions have the form: 

where n is a whole number.

Factor to find all the zeros x^3-5x^2+16x-30=0

factors of 30:

1,2,3,5,15,30
of which we'd determine that x=3 is a solution such that
x^3-5x^2+16x-30=(x-3)(x^2-2x+10)=0
but the quadratic is not factorable over the reals - so we'd need the quadratic formula to determine that x=1+3i, and x=1-3i are the other two factors. Thus the zeros are:
{3, 1-3i, 1+3i}

Find a cubic function with integral coefficients having (4 + 3i) and -3 as roots

If coefficients are real and roots are complex, then roots occur in complex conjugate pairs:

 
(x-(4+3i))(x-(4-3i))(x-(-3)) =
x^3-5x^2+x+75

How many zeros does the function g(x) = 4(x + 2)^5 (x-3)^3 + 5(x-3)^4 (x+2)^4 have?

the degree of this polynomial, when expanded is 8, so there are 8 complex solutions.