i need help with algebra factoring?

1. Find the difference: (8x2 + 6x + 3) - (4x2 + 3x + 2)

i got 9X²+8X+10?

The answer you gotten is totally wrong and I have no idea how you got that answer.

But I will show how to solve it as well the correct answer.

(8x^2 + 6x + 3) - (4x^2 + 3x + 2)

Write down the first group, but without the parenthesis. On the second group, distribute the negative 1. The negative 1 is located in front of the second group. It is known as an invisible 1.

8x^2 + 6x + 3 - 4x^2 - 3x - 2

Collect like terms and Factor

8x^2 - 4x^2 + 6x - 3x + 3 - 2

= 4x^2 + 3x + 1

The answer is 4x^2 + 3x + 1

2.Find the product: 2x(3x - 5)

i got 6X-10?

You got this wrong because you just distribute the 2 not 2x.

You need to distribute the 2x to each term.

2x(3x - 5x)

Like this:

(2x)(3x) - (2x)(5x)

Now multiply.

6x^2 - 10x^2

The answer is 6x^2 - 10x^2

3. Find the product: (2x + 1)(x + 3)

To solve use the FOIL method

(2x + 1)(x + 3)

(2x)(x) + (2x)(3) + (1)(x) + (1)(3)

Multiply each group.

2x^2 + 6x + x + 3

Combine like terms.

= 2x^2 + 7x + 3

The answer is 2x^2 + 7x + 3

4. Find the product: (x + 1)(x2 + x + 1)

i got x^2+3x+1

You got this wrong, because you just distribute x. You did not distribute the 1.

(x + 1)(x^2 + x + 1)

Distribute the x as well as the 1.

(x)(x^2) + (x)(x) + (x)(1) + (1)(x^2) + (1)(x) + (1)(1)

Multiply each term.

x^3 + x^2 + x + x^2 + x + 1

Collect like terms and add each like term.

x^3 + x^2 + x^2 + x + x + 1

= x^3 + 2x^2 + 2x + 1

The answer is x^3 + 2x^2 + 2x + 1

5.Find the product: (x + 2)(x - 2)

Use the FOIL method

(x + 2)(x - 2)

(x)(x) - (2)(x) + (2)(x) + (2)(-2)

Multiply each term.

x^2 - 2x + 2x + 4

Combine like terms

= x^2 + 4

The answer is x^2 + 4

6. Find the product: (x + 5)2

First thing is rewrite the problem

Instead of looking like this:

(x + 5)2

Make it look like this:

2(x + 5)

Distribute the 2 to each term

(2)(x) + (2)(5)

Multiply each term

= 2x + 10

The answer is 2x + 10

7. Factor the expression: x2 - x - 6

x^2 - x - 6

You need to think of 2 numbers that multiply to get - 6 at the same time get you - 1.

Those two numbers are: -3 and 2

So, write -3x and 2x

Then insert them into the problem itself to replace x

x^2 - 3x + 2x - 6

Group and Factor.

(x^2 - 3x) + (2x - 6)

x(x - 3) + 2(x - 3)

Group what is left on the outside of the parenthesis and write only one of whatever inside the parenthesis.

(x + 2)(x - 3)

The answer is (x + 2)(x - 3)

1) you can only add or subtract terms that have the identical variables and identical exponents

(8x^2 + 6x + 3 ) - (4x^2 + 3x + 2 ) = (8x^2 - 4x^2) + (6x - 3x) + (3 - 2) =

4x^2 + 3x + 1

2) you didn't include the variable x in the distribution

2x(3x - 5) = 2x (3x) - 2x (5) = 6x^2 - 10x

3) Use FOIL F- First term in each factor, O - outer terms in each factor, I - Inner terms in each factor and L - Last terms in each factor

(2x + 1)(x + 3) = 2x(x) + 2x(3) + 1(x) + 1 (3) = 2x^2 + 6x + x + 3 = 2x^2 + 7x + 3

4) same method as problem 3, just with another term

(x + 1)(x^2 + x + 1) = x(x^2) + x(x) + x(1) + 1(x^2) + 1(x) + 1(1) = x^3 + x^2 + x + x^2 + x + 1 =

x^3 + 2x^2 + 2x + 1

5 and 6 you should practice on your own.

5)x^2 - 4

6)2x + 10

7) look for the factor of the coefficient of x^2. In this case it is 1, and the factor pair is {1, 1}

look for the factor pairs to the constant term. In this case 6. the factor pairs are {1, 6} and {2, 3}

Notice that the sign in front of 6 is negative. that means the factors with have opposite signs

Is there a factor pair, that I can add or subtract to get a result of 1, 1 is the coefficient of x?

YES {2, 3} since the x term has a -1 , the 3 will be negative, and the 2 positive

(x + 2)(x -3)

Factoring take a lot of practice.

1. Not to be rude, but I have no idea where you got 9x^2. But you have to distribute the - into the second set of parentheses, then you combine like-terms (every x^2, x, and no variable). Should be 4x^2 + 3x + 1

2. Don't forget to multiply the X with the 2. Do each part inside the parentheses times 2X. Should be 6x^2 - 10x

3. Multiply each individual set of numbers. Use FOIL. First, Outer, Inner, Last. 2x*x, 2x*3, 1*x, 1*3, it should be 2x^2 + 7x + 3.

I'm going to stop here because this is really hard to type in. I hope you get how to do it. Also, make sure to remember that after Algebra, X will always be a variable and NEVER again will mean multiplication. So make sure you know that (x)(3)(4) is the same as saying x times 3 times 4, or 12x. Show multiplication with parentheses from now on.

1) (8x2 + 6x + 3) - (4x2 + 3x + 2)

Open brackets and group like terms

= 8x2 + 6x + 3 - 4x2 - 3x - 2)

= 8x2 - 4x2 +6x -3x +3 -2

= 4x2 +3x +1

2) 2x(3x - 5)

2x timex 3x - 2x times 5

= 6x2 - 10x

3. Find the product: (2x + 1)(x + 3)

Opening brackets

= 2x2 + 6x +x+3

= 2x2 + 7x +3

4. Find the product: (x + 1)(x2 + x + 1)

x(x2 +x +1) + 1(x2 + x +1)

= x3 + x2 +x + x2 +x+1

= x3 + 2x2 + 2x +1

5.Find the product: (x + 2)(x - 2)

x2 - 2x + 2x - 4

= x2 - 4

6. Find the product: (x + 5)2

2x + 10

7. Factor the expression: x2 - x - 6

(x-3)(x+2)

I can't understand what you did for #1. For question one, factor out the minus sign (treat it like you're multiplying whatever is in the parenthesis by -1) : 8x2 + 6x + 3 - 4x2 -3x - 2. Now just add/subtract the like terms (x2 and x2, x and x, no variable and no variable. You can't subtract x2 by x, or x2 by no variable, only by the same variable with the same exponent). So you should get 4x2 + 3x + 1.

#2, when you multiply, make sure you multiply x also. So x times x is x2, or times x times x is x3. So 2x(3x -5) you do (2x)(3x) - (2x)(5). That should give you 6x2 - 10x. Follow along with the rest of the problems.

1. All you have to do is subtract like terms:

8x^2 - 4x^2 = 4x^2

6x - 3x = 3x

3 - 2 = 1

So your answer is 4x^2 + 3x + 1

2. You must multiply 2x*3x and 2x*-5

The answer is 6x^2 -10x

3. Foil

2x^2 + 6x + x + 3 = 2x^2 + 7x + 3

4. x^3 + x^2 + x + x^2 + x + 1 =

x^3 + 2x^2 + 2x + 1

5. Foil

x^2 + 2x - 2x - 4 = x^2 -4

6. Foil (x+5)(x+5)

x^2 + 5x + 5x + 25 = x^2 + 10x + 25

7. (x - 3)(x + 2)

For a million) x^2 -x =12 You subtract 12 from both area to get: x^2 - x -12 =0 then you discover 2 numbers that when more desirable they supply your -12, yet even as extra, it resources -a million. hence, -4 and three. so that you position it in this format: (x-4)(x+3)=0 the reply is (x-4)(x-3) for 2) 2x^2 + 5x + 3 =0 (2x-3)(x+a million) for 3) 3x^2 + 2x -8 =0 (3x-4)(x+2) For 4) x^2 - 3x - 28 =0 (x+4)(x-7) for 5) 2x^2 - x -10=0 (2x-5)(x+2)

1) (8x2+6x+3)-(4x2+3x+2) 4) (x+1)(x2+x+1)

x2+x+1

x+1

_________

8x2+6x+3-4x2-3x-2 x3+x2+X

x2+x+1

4x2+3x+1=0 ____________

2) 2x(3x-5)

6x2-10x=0 x3+2x2+2x+1=0

3) ( 2x+1)(x+3) 5) (x+2)(x-2)= x2-4=0

2x+1

x+3 6) (x+5)2=x2+10x+25=0

----------

2x2+x 7) x2-x-6=(x-3)(x+2)=0

6x+3

---------------

2x2+7x+3=0