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I know how to do the ones where you find two factors that multiply together for the last number, and add to the middle # in the sequence, but I have no clue how to that up there and neither do my parents, so I would very much appreciate the help! Thanks so much!

2006-09-10 17:48:16 · 8 answers · asked by Anonymous in Science & Mathematics Mathematics

8 answers

Multiply 7 and 16.
7*16 = 112
Look for factors of 112 that add to -32
-28 times -4 = 112 and -28 + -4 = -32
Replace -32x with -28x - 4x
7x^2 - 28x - 4x + 16
Now factor by grouping
(7x^2 - 28x) + (-4x + 16)
7x^2 and 28x have 7x in common
and -4x and 16 have -4 in common
7x(x-4) + (-4)(x-4)
= (x-4)(7x-4)

2006-09-10 17:51:48 · answer #1 · answered by MsMath 7 · 2 0

(7x - 4) (x - 4)

This is the answer. Here are the general step.

1st step:
Determine the possible combination for the 7x^2 first. here are the list....

7 x 1
1 x 7 ....only

2nd Step :
Determine the possible combination for the +16.. Here are the possible list

-1 x -16
-2 x -8
-4 x -4
-8 x -2
-16 x -1
I put all to negative because i see a -32x in the center.

Third.. put them into the logical point and you will get the answer.

2006-09-11 00:56:49 · answer #2 · answered by Mr. Logic 3 · 0 0

This is a simple way.

Let the 7X^2 to 8x^2 -x^2

See:

7x^2 - 32x + 16 = 0
8x^2 - 32x - x^2 + 16 = 0
8x(x-4) - (x^2 - 16) = 0
8x(x-4) - (x-4)(x+4) = 0
(x-4)(8x - (x+4)) = 0
(x-4)(7x-4)=0

You could do that by solving the equation.

Be Succeed.

2006-09-11 00:57:52 · answer #3 · answered by Babax 3 · 0 0

Here is a guarantee everytime rather you can factor it or not

7x^2 - 32x + 16

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-(-32) ± sqrt((-32)^2 - 4(7)(16)))/(2(7))
x = (32 ± sqrt(1024 - 448))/14
x = (32 ± sqrt(576))/14
x = (32 ± 24)/14
x = (56/14) or (8/14)
x = 4 or (4/7)

So this would factor to

(x - 4)(7x - 4)

2006-09-11 01:53:16 · answer #4 · answered by Sherman81 6 · 0 0

they way i do it is:

first because its 7x^2 write (7x .....)(7x.....)/7

to find what goes in the spaces multiply 7 and 16 = 112

now you need two numbers which multiply to 112 and sum to -32

these numbers are -28 and -4 because -28*-4 = 112
and -28-4 = -32

therefore you get (7x -28)(7x -4)/7

this can be further simplified to get the final answer

(x - 4)(7x - 4).

2006-09-11 01:23:30 · answer #5 · answered by Anonymous · 0 0

7x^2 - 32x + 16

= (7x - 4)(x - 4)


Working,

We know the beginning is something like,

= (7x ......)(x.....) since this would give us 7x^2

We also know the ending is something like,

= (..... 4)(......4) since we need to get 16

Put them together,

= (7x ..... 4)(x ..... 4) and work out the signs

= (7x - 4)(x - 4), we get -28x - 4x = -32x and -4 x -4 = 16.

that's exactly what we need.

2006-09-11 01:15:01 · answer #6 · answered by ideaquest 7 · 0 0

find the discriminant of the following quadratic eqn
d=b^2-4*a*c
d=(32)*(32)-4*7*16
d=900+4+120-448
d=1024-448
d=576
let roots be r1&r2
r1=(-b+sqrt(d))/2a
r2=(-b-sqrt(d))/2a
r1=(32+24)/14
r2=(32-24)/14
r1=60/14
r2=8/14
thus
f(x)=(x-30/7)(x-4/7)

2006-09-11 01:10:20 · answer #7 · answered by sami1989 2 · 0 0

its trial and error to some respecct but here is the answer

(7x-4) (x-4)

2006-09-11 00:51:15 · answer #8 · answered by Anonymous · 0 0

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