Factorising Quadratics

[itaroken_]

I'm ever so crap at maths, and my cousin has asked for my help... So I've turned to the internet... (Sorry teachers) Any answer is appreciated, any response is awesome

 

Factorise these quadratics:

 

x² + 6x + 8 =

 

x² + 12x + 32 =

 

x² + 7x - 8 =

 

x² - 4x - 45 =

 

x² - 7x + 10 =

 

x² - 12x + 35 =

 

x² + 8x - 33 =

 

x² + x - 56 =

 

Thank you so much

 

 

[LukaP]

use the formula 

, where a, b and c are the numbers in the ax² + bx + c = 0 

 

to find the solutions for the equations. 

then when you have x₁ and x₂ you simply write the equation as (x - x₁)*(x - x₂)

[mansoor_]

use the formula 

 

to find the solutions for the equations. 

then when you have x₁ and x₂ you simply write the equation as (x + x₁)*(x + x₂)

**(x - x1)(x - x2)

[LukaP]

**(x - x1)(x - x2)

whoops indeed. typo

[CookiezFort]

Here is the first 4: I will do the otehr 4 when i have time

 

first one (x+4)(x+2)

 

second one is (x+8)(x+4)

 

third one (x+8)(x-1)

 

fourth one is (x+9)(x-5)

 

else do : X = -b +- square root of bsquared - 4ac all of that divided by 2a

 

A= The coeficient behind the X squared B = the number behind the X on the middle and C is the last number

[mansoor_]

whoops indeed. typo

I am surprised he didn't just type his question into wolfram alpha, would have taken the same time but he would have immediately got the answer. 

[LukaP]

I am surprised he didn't just type his question into wolfram alpha, would have taken the same time but he would have immediately got the answer. 

maybe he doesnt know about it.

 

OP, wolframAlpha is something you will like

[A.D.A.M]

use the formula

, where a, b and c are the numbers in the ax² + bx + c = 0

to find the solutions for the equations.

then when you have x₁ and x₂ you simply write the equation as (x - x₁)*(x - x₂)

love the formula too bad i cant use it all the time

[LukaP]

love the formula too bad i cant use it all the time

you can use it on any quadratic equation, as long as you move to the complex number plane, instead of the real number line as your solutions group. otherwise, no formula will let you calculate solutions to any quadratic

[Thunderjolt]

Watch this, It will help

[A.D.A.M]

you can use it on any quadratic equation, as long as you move to the complex number plane, instead of the real number line as your solutions group. otherwise, no formula will let you calculate solutions to any quadratic

no i meant they wont let me use it. sometimes theyll require me to complete the square or factorise

[LukaP]

no i meant they wont let me use it. sometimes theyll require me to complete the square or factorise

retarded school -.- use what is easy in math. thats why it exists. to make stuff easier.

[Mo5]

retarded school -.- use what is easy in math. thats why it exists. to make stuff easier.

"Your answer is correct, but you get 0 points because you didn't do it the way I showed it"

[itaroken_]

Oh my god I wasn't expecting that many replies

 

Thank you so much @LukaP @A.D.A.M @Thunderjolt @CookiezFort @mansoor_

 

I'll put in @CookiezFort answers and I'll take a look at the video, thanks so much!

 

And as for the formula, I don't think I'll be able to describe it to her, but I'll try

[A.D.A.M]

retarded school -.- use what is easy in math. thats why it exists. to make stuff easier.

*shrugs* its on the Cambridge syllabus iirc so they have to teach it.

[LukaP]

"Your answer is correct, but you get 0 points because you didn't do it the way I showed it"

i fucking hated that!!!!

 

Oh my god I wasn't expecting that many replies

 

Thank you so much @LukaP @A.D.A.M @Thunderjolt @CookiezFort @mansoor_

 

I'll put in @CookiezFort answers and I'll take a look at the video, thanks so much!

 

And as for the equation, I don't think I'll be able to describe it to her, but I'll try

anytime )

[Nineshadow]

These are soo soo easy...

 

Look :

x² + 6x + 8 = x^2 + 4x + 2x + 8 = x(x + 4) + 2(x+4) = (x+4)(x+2)

 

In general  we have :

ax^2 + bx + c

 

You need to split b into two numbers,let's call them z and y so : z + y = b and z * y = c

 

Here's an example :

x² + x - 56

 

b = 1

m + n = 1  (doesn't matter what you call them, you do this in your mind most of the time)

m * n = -56

 

Hm...7 * 8 = 56

So -7 * 8 = -56

-7 + 8 = 1

X^2 + 8x - 7x - 56 = x(x+8) - 7(x+8) = (x-7)(x+8)

 

Simple.

 

Of course, some quadratics are really hard to calculate like that and in those cases use the formulas @LukaP gave you.

[itaroken_]

Here is the first 4: I will do the otehr 4 when i have time

 

first one (x+4)(x+2)

 

second one is (x+8)(x+4)

 

third one (x+8)(x-1)

 

fourth one is (x+9)(x-5)

 

else do : X = -b +- square root of bsquared - 4ac all of that divided by 2a

 

A= The coeficient behind the X squared B = the number behind the X on the middle and C is the last number

 

So for x² - 4x - 45, what would the answer be?

 

Also, negative and negative is a positive right?

[Nineshadow]

So for x² - 4x - 45, what would the answer be?

 

Also, negative and negative is a positive right?

Yep.

 

Think about it.

I have 2 numbers.If I multiply them I get -45 and if I add them together I get -4...

Soo...they should be -9 and 5.

-9 * 5 = -45

5 - 9  = -4

x^2 - 9x + 5x - 45 = x(x-9) + 5(x-9) = (x+5)(x-9)

[itaroken_]

x

 

Yep.

 

Think about it.

I have 2 numbers.If I multiply them I get -45 and if I add them together I get -4...

Soo...they should be -9 and 5.

-9 * 5 = -45

5 - 9  = -4

x^2 - 9x + 5x - 45 = x(x-9) + 5(x-9) = (x+5)(x-9)

 Thanks

[itaroken_]

Yep.

 

Think about it.

I have 2 numbers.If I multiply them I get -45 and if I add them together I get -4...

Soo...they should be -9 and 5.

-9 * 5 = -45

5 - 9  = -4

x^2 - 9x + 5x - 45 = x(x-9) + 5(x-9) = (x+5)(x-9)

 

Sorry to bother you again, so I'm going to write a rule on her book and I'm wondering what the + and - equal...

E.G:

 

+ and - equals ?

 

- and + equals ?

 

- and - equals +

 

+ and + equals +

 

I always forget what those rules are

[Fate]

I'm ever so crap at maths, and my cousin has asked for my help... So I've turned to the internet... (Sorry teachers) Any answer is appreciated, any response is awesome

 

Factorise these quadratics:

 

x² + 6x + 8 =

 

 

x^2+6x+8

what are factors of 8 that also add up to 6?

4 and 2

4*2 = 8, 4+2 = 6

(x+2)(x+4)

[itaroken_]

x^2+6x+8

what are factors of 8 that also add up to 6?

4 and 2

4*2 = 8, 4+2 = 6

(x+2)(x+4)

Thanks

[Nineshadow]

Sorry to bother you again, so I'm going to write a rule on her book and I'm wondering what the + and - equal...

E.G:

 

+ and - equals ?

 

- and + equals ?

 

- and - equals +

 

+ and + equals +

 

+ and - is minus.

All those considering :

ax^2 + bx + c

b = m + n

We have a few cases :

If c is negative then m OR n must be negative.

If c is positive then  (m and n must be positive) OR (m and n must be negative)

[itaroken_]

+ and - is minus.

All those considering :

ax^2 + bx + c

b = m + n

We have a few cases :

If c is negative then m OR n must be negative.

If c is positive then  (m and n must be positive) OR (m and n must be negative)

Ah awesome, thanks for answering! You guys would have beaten my teacher so badly at explaining things