Factorize the following completely?

Question

(a) 2x^2-100


(b) 3x^2-8x+5

show working please

Answer 1

PART A:

Factor out the common 2.
2(x² - 50)

Unfortunately, 50 isn't a perfect square so if you are looking for integer factorization, you wouldn't factor further. The answer would be:
---> 2(x² - 50)

Note: if you do use the difference of squares, then you would have sqrt(50) = sqrt(25) * sqrt(2) = 5 sqrt(2). The final factoring would be:
---> 2 ( x + 5 sqrt(2) ) ( x - 5 sqrt(2) ) )


PART B:

The integer factors of 3 are 1 and 3, so start with:
(3x ... )(x ... )

Now the integer factors of 5 are:
1, 5 or -1, -5

You need a negative sign to get -8, so use -1 and -5.

Try them in either order. In one case you get -16x, the other gives -8x, so that is your answer:
---> (3x - 5)(x - 1)

Answer 2

2x^2-100
=2(x^2-50)
This cannot be factored further if rational numbers are required. If irrational numbers are allowed then
2(x^2 -50) = 2(x-5sqrt(2))( x+5sqrt(2))

3x^2 -8x +5
If this is can be factored, it will look like:
(3x - )(x - )
We know the above because it is the only way to get a first term that is 3x^2. We also know the middle ter m is negative so we can't get a negative term if we use +5 times +1 to get the required +5 for the last term . Therefore it must be -5 times -1 to give us the necessary +5 as the third term

Now the only question is where does the 1 go and where does the 5 go? Well, there are only two possibilities , so try one and see if it works. If it does, that's the answer If not try the other. If it works, that's the answer. If neither work, the expression cannot be factored.

I think you will quickly see that (3x-5)(x-1) is the correct answer because (-5)x + (-1)(3x) gives you the -8x middle term.

Answer 3

For a, first we need to factor out the two since 2x^2 and -100 are both divisible by 2.

a) 2(x^2-50)
Now we have a difference of two squares. x^2 is one of them and 50 is another. However, we can't find an exact square root of 50 (that is an integer). That doesn't matter, we can just take the square root of 50. So, a^2-b^2=(a+b)(a-b) so

2(x^2-50)=2(x-sq(50))(x+sq(50))

sq(50)=sq(25)*sq(2)=5sq(5)
Our final factored for is

2(x-5sq(2))(x+5sq(2))



b) 3x^2-8x+5
This is a tougher one. The process for these is a little more complicated. The goal is this:

We need to find two factors of 5, such that when one of them is multiplied by 3x and the other by x, the sum is -8x. The only factors of 5 are 5 and 1 (since 5 is prime). However, we notice that -8 is a negative, so we will need two negative factors of 5 (negative times negative is positive). So the factors are -1 and -5. We need to multiply one of them by 3x and the other by x to get -8x. We find that if we multiply -1 by 3x and -5 by x, the sum is -3x-5x=-8x. That's what we want! You need to use FOIL in reverse, and you get that the factored form is.
(3x-5)(x-1)

That's your answer. Hope this helps.

Answer 4

a) Factor out 2 first:
2(x^2 – 50)

You can use the Difference of Squares to further factor it:
2(x+SQRT(50))(x-SQRT(50))

You can factor out the SQRT(50) even further by recognizing that 50 has a square as a factor:
2(x+SQRT(50))(x- SQRT(50)) =
2(x+SQRT(25)*SQRT(2))(x- SQRT(25)*SQRT(2)) =
2(x+5*SQRT(2))(x-5*SQRT(2))

b) Foil the expression. You know that 3 and 1 are factors of 3. You know that 5 and 1 are factors of 5. So, your choices are pretty small:
3 * 5 + 1 * 1 = 16
3 * 1 + 1 * 5 = 8 < -- Bingo!
Since the last term is positive, you know that you are adding the factors together to get:
(3x – 5) (x – 1).

Answer 5

1) 2x^2-100
= 2(x^2-50)
= 2(x+5.2^1/2)(x-5.2^1/2)

2) 3x^2-8x+5
=3x^2 -3x-5x+5
= 3x(x-1) -5(x-1)
=(x-1)(3x-5)

Answer 6

a] 2x^2-100
=2[x^2-50]
=2[x+sqrt50][x-sqrt50]
b] 3x^2-8x+5
=3x^2-3x-5x+5
=3x[x-1]-5[x-1]
=[3x-5][x-1]


3x²+2x+5= answer

Answer 7

utilising the flow-examine: 2x + 8 | +8x x - 6 | -12x 2x^2 -40 8 | -4x you get (2x+8)(x-6)=0 subsequently (2x+8) =0 x-6 = 0 2x = -8 x=6 x = -4 utilising the quadratic formula, x= (4 ± ?sixteen +384) / 2(2) = 24/4 = 6 OR = -sixteen/4 = -4

Answer 8

a] 2x^2-100
=2[x^2-50]
=2[x+sqrt50][x-sqrt50]
b] 3x^2-8x+5
=3x^2-3x-5x+5
=3x[x-1]-5[x-1]
=[3x-5][x-1]

Answer 9

a.2x^2-100
2(x^2-50)
2(x+5rt2)(x-5rt2)

b.3x^2-5x-3x+5
=x(3x-5)-1(3x-5)
(3x-5)(x-1)