Algebra Factoring & Quadratic Equations?

Question

Could someone please help me with these. The " ^ " denotes an exponents.

-*- Quadractic Equations -*-
x^2 - 10 = 0
x^2 - 2X +1=0

-*- Factoring -*-
x^2 -7 x +10
5n^3 + 20n^3 +15n
81x^4 + 3x

Thank You!

Answer 1

x^2=10 x=+/- sqrt(10)

(x-1)^2=0 so x=1

You need two numbers that add up to -7 and multiply to 10 so -2 and -5
(x-2)(x-5)

5n^3+20n^3+15n
=25n^3+15n=5n(5n^2+3)

Of course if you meant 5n^3+20n^2+15n
=5n(n^2+4n+3)
=5n(n+1)(n+3)

3x(27x^3+1)
=3x(9x^2-3x+1)(3x+1) by sum of cubes formula

a^3+b^3=(a+b)(a^2-ab+b^2) where in the above equation a=3x and b=1

Answer 2

1) x^2 - 10 = 0

First, move the -10 to the right hand side.

x^2 = 10

Now, take the square root of both sides. Note that whenever you take the square root of both sides, you ALWAYS have to insert a +/- , or "plus or minus", on the right hand side.

x = +/- sqrt(10)

We cannot simplify sqrt(10) any further, so we just leave it. Therefore, your solutions for x are
x = {sqrt(10), -sqrt(10)}

2) x^2 - 2x + 1 = 0

Your first step is to factor this. This is a square binomial, as it factors into

(x - 1)(x - 1) = 0
(x - 1)^2 = 0

And now, take the square root of both sides. Normally, we would add a "+/-", but in this case, plus or minus 0 is the exact same thing, so it's not necessary.

x - 1 = 0, which means
x = 1.

Factoring:
3) x^2 - 7x + 10

Generally, when solving these (and whenever the coefficient of x^2 is 1, which, in this case, it is), you look for two numbers which add up to be the coefficient of x (-7) and two numbers which multiply to get +10. The answer in this case is -5 and -2; as a result, this factors into

(x - 5) (x - 2)

2) 5n^3 + 20n^2 + 15n

The first step would be to factor a monomial out of this; notice that common to all the terms is 5n. So, pull out a 5n, to get

5n(n^2 + 4n + 3)

Factoring this further using the same techniques we did before (two numbers that multiply to get 3 and add to get 4 is 3 and 1)

5n(n + 3) (n + 1)

3) 81x^4 + 3x

First, pull out the biggest monomial you can; in this case it's 3x.

3x (27x^3 + 1)

Now, you have what is called a sum of cubes, because 27x^3 is a cube and 1 is a cube (1 = 1^3). There is a methodology to factoring sums and differences of cubes.

Your first step would be, in the first set of brackets, to take the cube root of each term. For (27x^3 + 1), this would be:

(3x + 1) (? + ? + ?)

What goes in the question marks are the following steps, for sums and differences of cubes: "square the first" , "negative product" , "square the last". Remember these verbally.

I'll explain each of them.
a) "Square the first". What you must do is square the first term in the first set of brackets (3x + 1). As a result, we'll get (3x)^2 = 9x^2.

(3x + 1) (9x^2 + ? + ?)

b) "Negative product". Multiply the two terms in the first set of brackets together (i.e. take their product), and then multiply the result by -1 (i.e. take the negative). In our case, we'll have (3x)(1) which is equal to 3x, multiplied by -1 gives -3x.

(3x + 1) (9x^2 - 3x + ?)

c) "Square the last"

Square the second term. 1^2 = 1.

(3x + 1) (9x^2 - 3x + 1)

Now, our final answer [since we had a lingering 3x there to begin with], along with all the transition, is

81x^4 + 3x
3x (27x^3 + 1)
3x (3x + 1) (9x^2 - 3x + 1)

Answer 3

Hello :)

x^2 - 10 = 0
=> x^2 = 10
=> x = ±√10

x^2 - 2x + 1 = 0
=> (x - 1)(x - 1) = 0 (since (x - 1)^2 = x^2 - 2x + 1)
=> x = 1, 1

x^2 - 7x + 10
x^2 - 5x - 2x + 10
x(x - 5) - 2(x - 5)
(x - 2)(x - 5)

5n^3 + 20n^2 + 15n
5n(n^2 + 4n + 3)
5n(n^2 + n + 3n + 3)
5n[n(n + 1) + 3(n + 1)]
5n[(n +1)(n + 3)]
5n(n + 1)(n + 3)

81x^4 + 3x
3x(27x^3 + 1)
3x[(9x)^3 + 1^3]
3x(3x + 1)(9x^2 - 3x + 1)

Hope it helps :)

Answer 4

Leafsobsessed has it appropriate yet there's a small sign mistake. I’m borrowing her words and math with the correction blanketed. you go with for 2 factors that upload to the midsection form (10) and multiply to the 1st coefficient x the final form (3 x -8 = -24). so 12 and -2 paintings. you separate the midsection term, 10x, into 12x and -2x: 3x^2 +12x - 2x -8 then ingredient each a million/2 3x(x + 4) - 2(x+4) ingredient the x+4 out to get =(x+4) (3x-2) double verify via utilizing foil to get better the unique quadratic equation.

Answer 5

1. add 10 to both sides then take the square root
2. use the quadratic equation (or factoring which would be easiest)

3. (x-5)(x-2)
4. 5n(5n^2 + 3)
5. 3x(27x^3 + 1)= 3x(3x + 1)(9x^2 - 3x + 1)

Answer 6

a)x^2=10
x=+/-sqrt10
b)x^2-x-x+1=0
x(x-1)-1(x-1)=0
(x-1)^2=0
x=1
c)x^2-5x-2x+10
x(x-5)-2(x-5)
(x-5)(x-2)
d)5n(n^2+4n+3)
5n(n+3)(n+1)
e)3x(27x^3+1)

Answer 7

You don't need the Quadratic Equation for the first two.

the first answer is +- square root of 10
second answer is x=1