How do you factor $5 y^{2} - 2 y - 3$ $5y^2 - 2y - 3$ ?

Question

How do you factor $5 y^{2} - 2 y - 3$ $5y^2 - 2y - 3$ ?

2 Answers

Impact of this question

5379 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-12T00:42:47

$(5 y + 3) (y - 1)$ $(5y+3)(y-1)$

Explanation:

OK i'll try my best.

Think of a factorised equation as being in the form $(a y + b) (c y + d)$ $(ay+b)(cy+d)$

$a \times c$ $a xx c$ must equal $5$ $5$
$b \times d$ $bxxd$ must equal $- 3$ $-3$

So, what two integers multiply together to get 5? 5 and 1. So $a = 5$ $a=5$ and $c = 1$ $c=1$ So now you can write the equation as $(5 y + b) (y + d)$ $(5y+b)(y+d)$

What two integers multiply together to get -3? Well, there's four possibilities.
1: $b = 3 and d = - 1$ $b=3 and d=-1$
2: $b = - 3 and d = 1$ $b=-3 and d=1$
3: $b = 1 and d = - 3$ $b=1 and d=-3$
4: $b = - 1 and d = 3$ $b=-1 and d=3$

Which of these combinations gets you $5 y^{2} - 2 y - 3$ $5y^2-2y-3$ when you multiply out the brackets? Really, it's trial and error here, but it gets quicker as you do it more and more often. Combination 1 is the one that works.

$(5 y + 3) (y - 1)$ $(5y+3)(y-1)$

Answer 2 · 2018-04-12T00:47:48

Factor by grouping. You should get $(5 y + 3) (y - 1)$ $(5y+3)(y-1)$ at the end

Explanation:

Factor by grouping is by far the easiest factoring method I have ever encountered. First of all let me say that if you can factor a number out of the front number DO IT. Making the $x^{2}$ $x^2$ alone is so much easier to factor. In this case you cannot so let me should you my way.
Start by multiplying your $a$ $a$ term and $c$ $c$ term; if you do not know the base form of a quadratic equation is $a x^{2} + b x + c$ $ax^2 +bx +c$ :
When you multiply $5$ $5$ and $- 3$ $-3$ you get $- 15$ $-15$ . Now you need to find two numbers that multiply to $- 15$ $-15$ and add up to your $b$ $b$ term ( $- 2$ $-2$ ). In this case the two numbers are $- 5$ $-5$ and $3$ $3$ as you can see:
$- 5 + 3 = - 2$ $-5+3=-2$ and $- 5 \cdot 3 = - 15$ $-5*3=-15$ We're good to go.
Next step is to make the formula to factor:
Split you middle term into $- 5$ $-5$ and $+ 3$ $+3$ to make it true:
$5 y^{2} - 5 y + 3 y - 3$ $5y^2 -5y + 3y -3$
Next, put parenthesis around the first two variables and last two like so:
$(5 y^{2} - 5 y) (3 y - 3)$ $(5y^2-5y)(3y-3)$
Now this is starting to look like something you can factor. If you did everything right you should be able to factor the two parenthesis and get the same numbers inside both:
$5 y (y - 1) 3 (y - 1)$ $5y(y-1)3(y-1)$
If that is all right you can cross out one of the parenthesis and make a new one with the numbers you just factored:
$(5 y + 3) (y - 1)$ $(5y+3)(y-1)$
That's probably a little difficult to understand but I tried sorry.
To check just foil!!
$5 y^{2} - 5 y + 3 y - 3$ $5y^2-5y+3y-3$ checks out!!!

Science

Math

Humanities

... and beyond

How do you factor $5 y^{2} - 2 y - 3$ $5y^2 - 2y - 3$ ?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you factor 5y^2 - 2y - 35y2−2y−35y^2 - 2y - 3?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

How do you factor $5 y^{2} - 2 y - 3$ $5y^2 - 2y - 3$ ?