Factoring problems. Please help?

Question

Can you help me by solving these (or at least a few?) and show me your steps?


2) 49x^2 - 16y^2

4) 16 - x^4

6) 4x^2 + 9

7) 2x^2 - 5x - 3

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

Answer 1

2. difference of squares (7x + 4y)(7x - 4y)
4. same thing (4 + x^2)(4 - x^2) = (4 + x^2)(2 + x)(2 - x)
6. cannot be factored
7. basic factoring (x - 3)(2x + 1)
8. take out GCF x(3x^3 - 6x^2 - 4x + 8)

Answer 2

Hi,
The first three (2, 3, & 4) are the difference of two squares. This is a standard form that can be factored according to the equality:
(a² -b²) = (a - b) (a+b)
So, let's do the first one, and you can do the other two the same way.
1) Make two sets of parentheses and put the square roots of both terms in each. So, the square root of 49x² is 7x and the square root of 16y² is 4y.
(7x 4y) (7x 4y)
2) Now, put a minus sign in one binomial and a plus in the other. If doesn't matter which goes where.
(7x-4y)(7x+4y)
And that's all there is to it.
For the second one, you can write it as (x²)² and (2²)².

Now, let's do 7)
2x² -5x -3
Step 1: Write the factors of 2x² as the first term in two binomials:
(2x )(x )
Step 2: Now, find the factors of -3: {1, -3} and {-1, 3}.
Step 3: We want to choose the factors that will make the sum of the product of the inside and outside terms equal to -5x, the middle term of the original trinomail.
(2x +1)(x -3)
So, you see that 1*x + 2x *(-3) = x -6x = -5x.

Number 8:
This is considerabley more difficult:
Step 1: First factor out the common factor of x.
x(3x^3 -6x² -4x + 8)
Frankly, the best way to do this is with a graphing calculator. If you have a TI-83 Plus or TI-84, or even a TI-82, you can do it this way:
1) Press the Y= button and enter the expression into the y1 position. The "x" is on the [x,T,0,n] button.
2) Press GRAPH and the graph will be displayed.
3) Now, from the graph, it appears that one of the zeros is at x=2. So, press 2ND, CALC, ENTER, and then enter 2 and press ENTER again. You will get y =0, so (x-2) is one of the factors.
4) The other two factors are not whole numbers, so do them this way:
a) Press 2ND, CALC, 2 and move the cursor slightly to the left of the first point where the graph crosses the x-axis and press ENTER.
b) Move the cursor above the x-axis and press ENTER.
c) Press ENTER again and you will get x = -1.1547
Now, do those same steps with the second point where the graph crosses the x-axis.

If you don't have a calculator, you can do this:
The possible rational zeros are +- the factors of 8 divided by the factors of 3:
+- (1,2,4, 8)/(1,3). Each term in the numerator must be placed over each term in the denominator with +- signs.
Now, let's suppose that we made a lucky guess that 2 is a zero.
Let's do symthetic division to see.
2| 3 -6 -4 8
|___6_ 0_-8_
....3...0..-4....0
(The dots on the bottom row don't mean anything. They're just there to maintain spacing.)
So, since the last term is zero, it checks.
The coefficients of the reduced equation are the first numbers on the bottom row.
3x² -4 = 0
3x² = 4
x² = 4/3
x = +- 2sqrt(1/3)
I'm not sure if you wanted factors or zeros, but you can set them up as you wish.

Hope this helps.
FE

Answer 3

2) 49x^2-16x^2
=(7x+4y)(7x-4y)

4) 16-x^4
multiply that whole thing to make ur x positive
= x^4-16
=(x^2+4)(x^2-4)
the last set is still factorable so
=(x^2+4)(x+2)(x-2)
the reason the first part is not able to continue factoring is because you cannot factor it if there is a plus in between only 2 numbers you would have to have a 3rd number to be able to do so

6) 4x^2+9
this is NOT FACTORABLE . . . like I stated in the last one you must have a 3rd number in the middle to do so . . . therefore not factorable

7) 2x^2-5x-3
=(2x+1)(x-3)
negative 3 times positive 2 equals negative 6 and then u add a positive one and there is your negative 5 from the middle
***now unlike the last one this one does have a middle term . . .

8) 3x^4-6x^3-4x^2+8x
now when u have a 4 term to factor u split it up into 2 sets of () in the beginning like so:

=(3x^4-6x^3)*MINUS*(4x^2 *MINUS*8x)

***now as u see we left the minus in the middle of the 2 terms therefore to keep that 8x positive in the long run u must change it now if that 8x started out as a negative then u would have changed it to a positive

=3x^3(x-2)-4x(x-2)

**now this is factorable when u start off with 4 terms you want it to look something like this before u are finished, as you can see the () are the same which is exactly what you want so:

=(3x^3-4x)(x-2)

*u will take the parts that are not in () and put them together, and then u will put one set of the (x-2) . . . now for this to work the part that is left in the () must be the same if it is not then it wont work



Good luck

Answer 4

2) 49x^2 - 16y^2
= (7x)^2 - (4y)^2
= (7x-4y)(7x+4y)

4)16 - x^4
= 4^2 - (x^2)^2
= (4-x^2)(4+x^2)
= (2^2 - x^2)(4+x^2)
= (2-x)(2+x)(4+x^2)

* Note : 2 and 4 both use the formula A^2 - B^2 = (A-B)(A+B)

6) There is something wrong with the question. It cannot be factorised.

7) 2x^2 - 5x - 3
= 2x^2 - 2x - 3x -3
= 2x(x-1) - 3(x-1)
= (2x-3)(x-1)

8) 3x^4 - 6x^3 - 4x^2 + 8x
= x(3x^3 - 6x^2 - 4x + 8)
We notice if x=2 then the equation is equal to 0,
so the equation is divisible by (x-2)
using long division ( which i do not execute here due to inconvinience ) we have :
x(x-2)(3x^2-4)
Apply A^2 - B^2 = (A - B)(A + B) in 3x^2 - 4 we have :
x(x-2)(sqrt3x - 2)(sqrt3x + 2)
That is the factorised equation

Answer 5

4) 16 - x^4
16 = x^4
2^4 = x^4
x = 2

6) 4x^2 + 9
4 x^2 = -9
x^2 = -9/4
x = negative square root 9/4
x = -3/2

7) 2x^2 - 5x - 3

= (2x +1)(x - 3)
** it's a little hard for me to show the method of solving this because I'd have to explain it.

-- I'm working on the rest.

Answer 6

2) 49x^2 - 16 y^2

= 7^2x^2 - 4^2y^2

implies (7x-4y)(7x+4y)


4) 16- x^4

= 2^4-x^4 = (2^2-x^2)(2+x^2)


6) 4x^2+9 = 2^2x^2 + 3^2 = 4 ( x^2 +3^2)



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

Answer 7

#2 and #4 are difference of squares questions

#6 does not factor

#7 factors to (2x +1)(x -3)

#8 factors by GCF
x(3x^3 -6x^2 -4x +1)

Answer 8

2) 49x^2-16y^2= (7x)^2-(4y)^2 =(7x+4y)(7x-4y)Ans

4) 16-x^4 =4^2-(x^2)^2=(4+x^2)(4-x^2)=(4+x^2)(2^2-x^2)
=(4+x^2)((2+x)(2-X)Ans

Answer 9

2) (7x-4y)(7x+4y)

4) (16-x^2)(16+x^2)