Factorize help again!!?

Question

I really don't get how to factorise something, my teacher doesn't explain things properly, could someone tell me how to work out these 3 questions?
1.)5x^2-19x+12
2.)3x^2-11x-20
3.)4x^2+17x-15
can someone show me step by step how to factorise them with the answer please because this is driving me crazy!

Answer 1

1) 5x^2-19x+12

5x^2 * 12 = 60x^2

factors of 60x^2

2x,30,
3x,20x
4x,15x

so -15 - 4x = -19x (the middle term of the equation
and -15x * - 4x = + 60x^2 ( the term got from multiplying the two extreme terms of the equation i.e.,5x^2 * 12 = 60x^2

so
the equation becomes:

5x^2 - 15x - 4x + 12 = 0
5x(x - 3) - 4(x - 3) = 0
(5x - 4)(x-3) = 0

5x - 4 = 0
5x = 4
x = 4/5

or
x - 3 = 0
x = 3

therefore, x = 4/5, 3


2) similarly 3x^2-11x-20 = 0
3x^2 - 15x + 4x - 20 = 0
3x(x - 5) + 4(x - 5) = 0
(3x + 4)(x-5) = 0

3x + 4 = 0
x = -4/3

or
x -5 = 0
x = 5

3) 4x^2+17x-15 = 0
4x^2 + 20x - 3x - 15 = 0
4x(x + 5) - 3(x + 5) = 0
(4x - 3)(x +5) = 0

4x - 3 = 0
x = 3/4

or
x + 5 = 0
x = -5

Answer 2

Factoring by Grouping

Step 1 Write the quadratic expression in standard form: ax^2 + bx + c = 0

Step 2 Write a table of pairs of factors of the product of the leading coefficient and the constant or the product ac.

Step 3 Choose the pair of factors that have a sum equal to the coefficient of the middle term. That is, find the pair of factors of ac that have a sum of b.

Step 4 Replace the middle term with two terms that use the factors from step 3 as coefficients.

Step 5 Factor the first two terms and the last two terms. Then, factor out the common binomial factor.

1) 5x^2-19x+12
= 5x^2-15x-4x+12
= 5x(x-3)-4(x-3)
= (5x-4)(x-3)

2) 3x^2-11x-20
= 3x^2-15x+4x-20
=3x(x-5)+4(x-5)
=(3x+4)(x-5)

3) 4x^2+17x-15
= 4x^2+20x-3x-15
=4x(x+5)-3(x+5)
=(4x -3)(x+5)

Some people use the quadratic formula to factor quadratic expressions:

Given 5x^2-19x+12. Consider the equation 5x^2-19x+12=0 where a = 5, b = -19, and c = 12.

x = [-b +or- sqrt(b^2 - 4ac)] / (2a)

x = [-(-19) +or- sqrt((-19)^2 - 4*5*12)] / (2*5)

x = [19 +or- sqrt(121)] / 10

x = (19 + 11) / 10 or x = (19 - 11) / 10

x = 3 or x = 4 / 5

Each root, let us call it m, has a related factor, x - m, of the original equation.

x = 3 or x = 4 / 5 im[plies
(x - 3)(x - 4/5) = 0
(x - 3)5(x - 4/5) = 5*0
(x - 3)(5x - 4) = 0

So, the factored form of 5x^2-19x+12 is (5x - 4)(x - 3).

Answer 3

ok
1) 5x^2-19x+12
(5x - ) (x - ) you can get that much because the numbers in the first spot must multiply to get 5x^2 and since 5 and 1 are the only things that work you have that. and here is a trick that ALWAYS works the + in front of the 12 means that the signs when you factor will be the same and the - tells you that they will both be a - sign.
now the #s in the last spots must multiply to get 12 and add to -19( the -19 after you multiply a factor by 5)
so through guess and check you get
(5x-4)(x-3) FOIL them together to make sure and you get 5x^2-19x+12

2) (3x+4)(x-5) that - in front of the 20 means that the signs must be different.. so these are a bit harder
3)(4x-3)(x+5)

Answer 4

its a bit confusing so hope i can explain properly..

1. first u need to determine the factors of the first and last term: eg. first term: 5 X 1 last term: 12 x 1, 2 x 6, 4 x 3

2. write down the factor of the first term vertically and the factor of the last term vertically next to it, like so:

5 4
1 3
its gona take some practise deciding on the correct factor to use with the last term- but with practice it gets very easy

3. you need to multiply the values diagonally (imagine a cross connecting the 5 and 3, and the 1 and 4)
5 4 = 4
1 3 = 15

4. k now the two answers (the 4 and 15) combined need to give you the value of the middle term in the question (-19 in this case) therefore both terms are negative (-15 - 4 = -19)

5. reading horizontally and including the - (we just found that both are -) we get:

(5 x - 4)( x - 3)
thats the final answer

the answers for the other two are:
2) (3x + 4)( x - 5)
3) (4x - 3)(x + 5)

see if you can get the answers for question 2 and 3 using this method. hope it helps

Answer 5

1.) 5x^2 - 19x + 12
The only factors for 5 are 5 and 1, since 5 is prime so we know that one of the two factors will be 5x and the other will be x.

(5x +?)(x+?)

The possible factor pairs of 12 are
(12,1), (6,2), and (4,3)

The easiest thing to do now is try the pairs, and with experience you'll get a feel for which one is the most likely pair. Remember when you do try them that they could go in either order, meaning you'd have to try BOTH (5x+12)(x+1) AND (5x+1)(x+12)

Experience, though tells me that the most likely pair is the 3 and 4.

(5x+4)(x+3) is the correct factoring since
(5)(3)+4 = 19


2.) 3x^2 - 11x - 20
Again, the 3 is prime so we know that the factors must be
(3x +?)(x+?)
The factors of -20 are
(20, -1), (-20,1), (10,-2), (-2,10), (5,-4), (-5,4) (and remember they can go in the opposite order)

Again with trial and error, or using experience as a guide, we'll find that the correct pairing is
(3x + 4)(x-5) since
(3)(-5) + 4 = -11

3.) 4x^2 + 17x -15
This one is potentially trickier because 4 is not prime. The factors could be (4,1) or (2,2).

So we have 2 possible candidates for the factoring of the first term:
(4x+?)(x+?)
or
(2x+?)(2x+?)
BUT we know that 2 times any number plus 2 times any other number is going to be an EVEN number. The sum of these is actually -17, which is odd, so we know that we must use the first expression.

(4x+?)(x+?)
Factors of -15 are
(15,-1), (-15,1), (5,-3), and (-5,3) in either order

Trial and error gives us
(4x+3)(x-5) since
(4)(-5)+3=-17

Answer 6

It's 99% understanding how to multiply binomials.
5x² - 19x + 12 will factor into
(5x + ?)*(x + ??). How do I know this? Becasue 5 is a prime number. How do I get the ? and ?? parts? I know that their product is +12 so they both have to have the same sign, and I know that sign has to be negative because the 'middle' term is negative. Now, 12 factors into 1*12, 2*6, or 3*4 and I need to be able to multiply one of those factors by 1 and the other by 5 and have their sum be 19. Well..... 3*4 looks like a good choice because 5*3 is 15 and adding 4 makes 19. And they both have to be negative, so.......
(5x - 4)*(x - 3) and then multiply it out.
5x*x -4*x - 3*54 (-3)*(-4) = 5x²-19x + 12 YAY!! ☺
Now --you-- do the rest of them because you won't ever get to be any good at it if you don't practice. ☺

Doug

Answer 7

There are several ways. Are you good at Algebra?
think of ax^2+bx+c
1) 5x^2-19x+12
a=5, b=-19 c=12
calculate, [-b+sqrt(b^2-4ac)]/2a ( the whole thing should be divided by 2a)
= [-(-19)+sqrt(19^2-4(5)(12)]/(2x5)
=[19+sqrt(121)]/10
=[19+11]/10
=30/10
=3 we have one answer
calculate [-b-sqrt(b^2-4ac)]/2a
= [-(-19)-sqrt(121)]/10
=19-11/10
=8/10
so, the final answer (x+3)(x+8x/10)
=(x+3)(x+4x/5)
Can you try the others? Use the same principle.

Answer 8

You can also use the formula x = -b +/- sqrt (b^2-4ac) / 2a.

1. x = [19 +/- sqrt(121)] / 10 = [19 +/- 11] / 10 = 3 or 4/5.
2. x = [11 +/- sqrt(361)] / 6 = [11 +/- 19] / 6 = 5 or -4/3.
3. x = [-17 +/- sqrt(529)] / 8 = [-17 +/- 23] / 8 = -5 or 3/4.

So to factor:

1. (x-3)(5x-4)
2. (x-5)(3x+4)
3. (x+5)(4x-3)

Answer 9

1)

5x² - 19x + 12 = 0

5x² - 15x - 4x + 12 = 0

5x(x - 3) - 4(x - 3) = 0

(5x - 4)((x - 3) = 0

- - - - - - - - - -

2)

3x² - 11x - 20 = 0

3x² - 15x + 4x = - 20 = 0

3x(x - 5) + 4(x - 5) = 0

(3x + 4)(x - 5) = 0

- - - - - - - -

3)

4x² + 17x - 15 = 0

4x² + 20x - 3x - 15 = 0

4x(x + 5) - 3(x + 5) = 0

(4x - 3)(x + 5) = 0

- - - - - - - - s-

Answer 10

1) 5x^2-19x+12
= 5x^2-15x-4x+12
= 5x(x-3)-4(x-3)
= (5x-4)(x-3)

2) 3x^2-11x-20
= 3x^2-15x+4x-20
=3x(x-5)+4(x-5)
=(3x+4)(x-5)

3) 4x^2+17x-15
= 4x^2+20x-3x-15
=4x(x+5)-(3x+5)
=(x+5)(4x-1)