HELP!!! I really suck at factoring!!?

Question

How do u factor these two expressions:

5x^2-14x+8

and 3x^2-7x-20?

Answer 1

Question 1
Factors of 5x² must be 5x and x
(5x"""""")(x""""")
Factors of 8 could be 4 and 2
(5x""""""4)(x""""""2)
Now allocate signs to obtain +8 and -14x
(5x - 4)(x - 2)

Question 2
Method as above:-
(3x + 5)(x - 4)

Answer 2

5x^2 - 14x + 8
multiply the coefficient of x^2 and the constant in the equation 8 you get: 5 x 8 = 40. Find factors of 40(-10 x -4) such that when added together will give you -14. Then substitute these factors in place of -14 in the equation to break the equation up for factoring. The equation becomes:
5x^2 - 10x - 4x + 8
Group the eqations and factor out common terms:
5x(x - 2) -4(x - 2)
Since x - 2 is common to both brackets, the factored equation will be:
(5x - 4)(x - 2)


Using the same reasoning as in the above equation we have:
3x^2 - 7x - 20
3 x -20 = -60((-12, +5) are factors of -60 such that when added together will give -7). We substitute these for -7:
3x^2 - 12x + 5x - 20
3x(x - 4) + 5(x - 4)
The factored equation becomes:
(3x + 5)(x - 4)

Answer 3

Just start guessing and then multiply out (foil) until you get the correct factors.

For example 5x^2 comes from 5x times x so you can write

(5x )(x ) to get started.

Now go to the last term 8. This comes from multiplying either 1 x 8, 8 x 1, 4 x 2, or 2 x 4 in the second position of each parenthesis.

So the possibilities are:

(5x-1)(x - 8), (5x - 8)(x-1), ( 5x-4)(x-2) or (5x -2)(x-4)

We use negatives in both () since the product must be +8 but the middle term negative.

Now multiply out each pair above until you get xx^2-14x+8.

Remember, using the distributive property to multiply can be remembered by FOIL which mean multiply First terms, the the Outside and Inside terms and add the results, and finally the Last terms.

By my calculation the third pair of factors are the ones that work.

Now try the same method on the second problem. This time though, since the last term is -20, the last terms must be of opposite signs.

Answer 4

Problem 1:
5x^2 - 14x + 8 = 0
(5x - 4) (x - 2) = 0
5x - 4 = 0, 5x = 4, x = 4/5
x - 2 = 0, x = 2

Answer: x = 4/5 or 2

Proofs:
= 5(0.8^2) - 14(0.8) + 8
= 3.2 - 11.2 + 8
= 0

= 5(2^2) - 14(2) + 8
= 20 - 28 + 8
= 0

Problem 2:
3x^2 - 7x - 20 = 0
(3x + 5) (x - 4)
3x + 5 = 0, 3x = - 5, x = - 5/3
x - 4 = 0, x = 4

Answer: x = - 5/3 or 4

Proofs:
3(- 5/3^2) - 7(- 5/3) - 20 = 0
3(25/9) + 35/3 - 20 = 0
25/3 + 35/3 - 20 = 0
20 - 20 = 0

3(4^2) - 7(4) - 20 = 0
48 - 28 - 20 = 0

Answer 5

to factor each expression, find the factors of the coefficient of the quadratic [x^2] and the constant [x^0] terms. choose pairs of factors of the quadratic and constant coefficient that sum to the coefficient of the linear term.

for the first trinomial,
5x^2-14x+8
+-1 and +-5 are factors of the coefficient of the quadratic term
+-1,+-2,+-4,+-8 are factors of the linear term

find the factors for the second trinomial and work from there.

Answer 6

5x^2-14x+8
= (5x-4)(x-2)

3x^2-7x-20
= (3x+5)(x-4)

Answer 7

1. 5x^2 - 14x + 8

5x^2 - 10 x - 4x + 8

5x(x-2) - 4(x-2)

(5x-2)(x-2)

2. 3x^2 - 7x - 20

3x^2 -12x + 5x - 20

3x(x-4) + 5(x-4)

(3x+5)(x-4)

Answer 8

(5x - 4)(x - 2)

(3x + 5)(x - 4)