Help me understand?

Question

Complete the following statement. x2 + 13x + 42 = (x + 6)( )

Answer 1

Hi,

Suppose you were only given x² + 13x + 42, and you had to factor it all on your own.

I' m going to explain factoring trinomials the way that I teach it, which is probably different from what you've done. Try to follow the steps. This method will work on any trinomial.

x² +13x + 42 Look for a GCF None this time. If there was one, factor it out. Then temporarily start both parentheses
............................. with the first number and variable.
(x.......)(x..........) First sign goes in first parentheses.
(x..+....)(x..........) Product of signs goes in 2nd parentheses.
(x..+....)(x...+.....) <== plus is because pos x pos = positive

Now multiply your first and third numbers together. Ignore their signs -> you've already done them. 1 x 42 = 42 So, out to the side list pairs of factors of 42.

42
------
1, 42
2, 21
3,14
6,7

Now you want to pick which factors go in your parentheses, using these rules:

If the signs you put on your parentheses are the SAME, find the factors that ADD up to the middle number in the problem.
If the signs you put on your parentheses are the DIFFERENT, find the factors that SUBTRACT to the middle number in the problem. (Note you look at signs in the parentheses, not in the problem.)

(x..+....)(x...+.....) Your signs are the same, so you want to add factors to get 13. Those factors are 6 and 7. When you put the numbers in the parentheses, the bigger number is pushy - it always goes first. So your factors now are:
(x.+.7)(x.+.6)

Those are your factors, so this shows that (x + 7) is the missing factor.

Here are some more examples of this factoring method:

3y² - 13y - 10 Look for a GCF None this time. If there was one, factor it out. Then temporarily start both parentheses with the first number and variable.
(3y.......)(3y..........) First sign goes in first parentheses.
(3y..-....)(3y..........) Product of signs goes in 2nd parentheses.
(3y..-....)(3y...+.....) <== plus is because neg x neg = positive

Now multiply your first and third numbers together. Ignore their signs - you've already done them. 3 x 10 = 30 So, out to the side list pairs of factors of 30.

30
------
1, 30
2, 15
3, 10
5, 6

Now you want to pick which factors go in your parentheses, using these rules:

If the signs you put on your parentheses are the SAME, find the factors that ADD up to the middle number in the problem.
If the signs you put on your parentheses are the DIFFERENT, find the factors that SUBTRACT to the middle number in the problem. (Note you look at signs in the parentheses, not in the problem.)

(3y..-....)(3y...+.....) Your signs are different, so you want to subtract factors to get 13. Those factors are 2 and 15. ( Notice that 3 and 10 would have added to 13, so you HAD to know to subtract this time.) When you put the numbers in the parentheses, the bigger number is pushy - it always goes first. So your factors now are:
(3y..-..15)(3y.+.2.)

Now you have to reduce either or both parentheses by dividing each parentheses' terms by the largest possible divisor. In our problem, the first parentheses is divisible by 3 but the second parentheses does not reduce.

(3y..-..15)(3y.+.2.)
-------------
.......3 This reduces to your final factors of

(y - 5)(3y + 2)

************************
NEXT PROBLEM !!

2x² +15x + 7 Look for a GCF None this time. If there was one, factor it out. Then temporarily start both parentheses
............................. with the first number and variable.
(2x.......)(2x..........) First sign goes in first parentheses.
(2x..+....)(2x..........) Product of signs goes in 2nd parentheses.
(2x..+....)(2x...+.....) <== plus is because pos x pos = positive

Now multiply your first and third numbers together. Ignore their signs - you've already done them. 2 x 7 = 14 So, out to the side list pairs of factors of 14.

14
------
1, 14
2, 7

Now you want to pick which factors go in your parentheses, using these rules:

If the signs you put on your parentheses are the SAME, find the factors that ADD up to the middle number in the problem.
If the signs you put on your parentheses are the DIFFERENT, find the factors that SUBTRACT to the middle number in the problem. (Note you look at signs in the parentheses, not in the problem.)

(2x..+....)(2x...+.....) Your signs are the same, so you want to add factors to get 15. Those factors are 1 and 14. When you put the numbers in the parentheses, the bigger number is pushy - it always goes first. So your factors now are:
(2x.+.14)(2x.+.1)

Now you have to reduce either or both parentheses by dividing each parentheses' terms by the largest possible divisor. In our problem, the first parentheses is divisible by 2 but the second parentheses does not reduce.

(2x.+.14)(2x.+.1)
-------------
.......2 This reduces to your final factors of

(x + 7)(2x + 1)

I hope that helps you!! :-)

Answer 2

write like this: x2+13x+42 = (x+6)(x+a) (since you know that x*? becomes x2)

Then solve (x+6)(x+a): x2+ax+6x+6a
Then compare with the first equation: x2=x2
ax+6x=13x -> x(a+6)=x(13)-> a+6=13 ->a=7
6a=42 ->a=42/6 -> a=7

that means: x2+13x+42 = (x+6)(x+7)

Answer 3

x² + 13x + 42 =

x² + 7x + 6x + 42 = 0

Group Factor

(x² + 7x) + (6x + 42)

x(x + 7) + 6(x + 7) = 0

(x + 6+(x + 7) = 0

- - - - - - - -s-

Answer 4

(x+7)

When you multiply (x+6)(x+7) you foil it out to get
x^2 + 13x + 42

Answer 5

(x+7) is the answer

x^2 + 13x + 42
x^2+7x+6x+42
x(x+7)+6(x+7)
(x+6)(x+7)