How do you "solve the quadratic equation for x by factoring"?

Question

My math packet asks me to solve the quadratic equation for x by factoring and i don't know how.

Here are some of the questions:
1. 6x^2 + 3x = 0
2. x^2 - 2x - 8 = 0
3. x^2 + 10x + 25 = 0

Can someone explain this to me, thanks.

Answer 1

Question 1
3x (2x + 1) = 0
x = 0 , x = - 1 / 2

Question 2
(x - 4) (x + 2) = 0
x = 4 , x = - 2

Question 3
(x + 5) (x + 5) = 0
x = - 5 (twice)

Answer 2

Question 1

6x^2 + 3x = 0 ==> 3x (2x + 1) = 0

Now you need to figure out the two values of x that make the left hand side of the equation 0.

The first obvious answer is 0.

Now you need to work out:

2x + 1 = 0 ==> 2x = -1 ==> x = -1/2

Therefore, x1 = 0, x2 = -1/2

Q.E.D.


Question 2

x^2 – 2x – 8 = 0

Add 1 to both sides of the equation:

x^2 – 2x + 1 – 8 = 1

Add 8 to both sides of the equation:

x^2 – 2x + 1 = 9 ==> (x – 1)^2 = 9

Now find two numbers that make (x - 1)^2 equal to 9
Therefore, x1 = - 2, x2 = 4

Q.E.D.



Question 3

x^2 + 10x + 25 = 0 ==> (x + 5)^2 = 0

Therefore, x1 = x2 = -5

Q.E.D.

Answer 3

the first one doesn't require factorising.
6x^2= -3x
cacelling x from both sides

6x=-3

x=-3/6=-1/2

now,in actorsing method,what we try to do is that,when a quadratic equation is given,
ax^2+bx+c=0,
find two numbers such that their sum is b and product is c.
as an example,
x^2 -2x -8=0

numbers whose product is -8 are:
-1 X8
-2X4
-4 X2
-8X1

once u take a closer look, u can see that -2 +4=2 which is our b term.

so the next step is,
u write the equation as
[x-2][x+4]=0
this means that either [x-2]=0 or [x +4]=0
so,x=+2,-4.

in the next one,
x^2 +10x +25=0
numbers whose product is 25:
1X25
5X5
u can see that,5 +5=10.
so our numbers are 5,5

so,we rewrite the equation as:
[x+5][x+5]=0
so,x=-5,-5.

i woulnd't suggest this method because once u get to higher classes,all quad, equations are too complex to be solved like this.use the quadratic formula.

Answer 4

1.6x^2+3x=0
3x(2x+1)=0
the product of two terms/expressions can be zero only when one of the terms/expressions is zero.Hence we can say that either 3x=0 or 2x+1=0
If 3x=0,then x=0
if 2x+1=0,then 2x= -1 or x= -1/2
therefore x=0 or -1/2 ans
2.x^2-2x-8=0
or,x^2-4x+2x-8=0
or,x(x-4)+2(x-4)=0
or,(x-4)(x+2)=0
therefore either x-4=0 or x+2=0
If x-4=0,then x=4
if x+2=0,then x= -2
x= 4 or -2 ans
3.x^2+10x+25=0
or (x)^2+2*x*5+(5)^2=0
or,(x+5)^2=0
or,x+5=0
x= -5 ans
.

Answer 5

Haha
Hey are you going to be in Mr.Meloys Precalc class tomorrow? I have the exact same packet if so and I can answer any more questions you got.