How do you factorise this 6xSquared + 36x +30?

Question

Into two brackets

Answer 1

6x^2+36x+30
=6(x^2+6x+5)
=6(x^2+x+5x+5)
=6{x(x+1)+5(x+1)}
=6(x+1)(x+5)

Answer 2

6x^2 + 36x + 30
30 x 6 = 180 <--this is a x c, so i break up 36x so the product of the numbers equals 180. 6 and 30 add to 36 and multiply to 180.

(6x^2 + 6x) + (30x + 30), factor each one
6x(x + 1) + 30(x + 1)
(6x + 30)(x + 1)

Answer 3

6x^2 + 36x + 30

First: factor - find the least common factor, which is 6.

6(x^2 + 6x + 5)

Sec: factor x^2+6x+5...multiply the 1st & 3rd term to get 5. find two numbers that give you 5 when multiplied & 6 (2nd term) when added/subtracted. the numbers are 1 & 5. rewrite the expression with the new middle terms.

x^2 + 1x + 5x + 5

*GRoup "like" terms & factor both sets of parenthesis.

(x^2 + 1x) + (5x + 5)
x(x+1) + 5(x+1)

Third: make sure the terms in parenthesis are the same - combine the inner term (once) with the outer term, along with the least common factor.

6(x+1)(x+ 5), or 6(x+5)(x+1)

Answer 4

6x^2 + 36x + 30
first multiply the first and tird number ( 6 x 30 = 180)
then w have to find 2 nmber that if we multiply it,it will prodce180 and if we add the two number it will produce 36(the second number)

like this :

_ x _ = 180
_ + _ = 36

the two nmber must be 30 and 6

6 x 30 = 180
6 + 30 = 36

we change the number 36x with these two number
36x = 6x +30x

then 6x^2 + 36x + 30
6x^2 + 6x + 30x + 30
6x(x + 1) + 30(x + 1)
(6x + 30) (x + 1)

Answer 5

6 (x ² + 6 x + 5)
6 (x + 5) (x + 1)

Answer 6

First of all factor out 6 & you will have
6(xsquared+6x+5),then factorise what is in the brackets
6(xsquared+5x+x+5)
6[x(x+5)+1(x+5)]
6[(x+5)(x+1)]

Answer 7

Step 1: Factor out 6 in each term by dividing by 6 by each term:

6(x^2 +6x + 5)

Step 2: Since 6 is just a plain number set it aside and just work with x^2 +6x + 5. Your answer will not change. Start by drawing two parenthesis and think about what you know...

( )( )

Step 3:Look at x^2... What times what will give you x^2. Only x times x with work so place an x at the beginning of each parenthesis...

(x )(x )

Step 4: Next look at the +5 at the end of the equation. Think about what times what gives you 5. Only 1 times 5 will work. So place a one and a 5 at the end of each parenthesis.

(x 1)(x 5)

Step 5: The last step is to figure out what sign goes inside each parenthesis. To do this reason what will get you a +6x and +5. The easiest thing to do is multiply the answer out leaving out signs... You can then reason logically what the signs must be.

(x 1)(x 5)

FOIL
F = First multiply the first terms... you get x times x = x^2
O = Second multiply outer terms... you get x times 5 = 5x
I = Third multiply inner terms... you get 1 times x = 1x
L = Second multiply outer terms... you get x times 5 = 5

So overall you get:
x^2 5x x 5

You want 5 to be positive... Your two options to get 5 are:
5 x 1 OR
-5 x -1

In order to get a +6x... 5x and 1x must be positive.. Therefore, 5 and 1 must each be positive...

That lease us with:

(x + 1)(x + 5)

Finally, a 6 was factored out earlier, so the final answer is:

6(x + 1)(x + 5)

Answer 8

6(x^2 + 6x + 5) = 6(x+5)(x+1)

Answer 9

6x^2 + 4x - 36 = 0 Use the quadratic fixing formula x = [ -b +/- SQRT(b^2 - 4ac) ] / 2a a=6, b=4, c=-36 So x = [ -4 +/- SQRT(sixteen + 864) ] / 12 = [-4 +/- sq.(880) ] / 12 = [-a million +/- sq.(a hundred and ten) ] / 3 One root is x=[sq.(a hundred and ten) - a million] / 3 the different root is x= - [sq.(a hundred and ten) + a million] / 3 Factorised: 6x^2 + 4x - 36 = 6 * ( x - [sq.(a hundred and ten)-a million]/3 ) * ( x + [sq.(a hundred and ten)+a million]/3 )

Answer 10

(6x + 6)(x + 5)