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Factoring

+1
199
12
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We were given factoring questions and no one ive contacted has any clue how to do them so to the internet i go!

1) 3a^2-10a-8

and

2) 30x^2-9x-3

and

3) 16-25x

also

4) 4(2-a)^2 -81

thanks in advance :)

Nov 21, 2018

#1
+18329
+2

Example

1   (3a +2)(a-4)            the  numbers   2   and -4   must multiply to equal  '-8'

2    (15x+3)( 2x-1)        again.....the   3  and  -1    multiply to -3   and the 15 and 2 multiply to 30

3     This one is factored as far as it can be

4    4 (a^2 -4a +4) - 81

4a^2-16a+16 - 81

4 a^2-16a-65

Nov 21, 2018
#4
+702
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You can simplify $$4a^2-16a-65$$

CoolStuffYT  Nov 21, 2018
#5
+18329
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If you use the quadratic formula you will find   a= 52/8   or -20/8

then you would have      (a-52/8)(a+20/8)        but you will see that   a x a = a^2   NOT 4 a^2.....

so lets multiply both terms by 2

(2a-52/8 * 2)( 2a+20/8 *2)

(2a-13)(2a+5)

ElectricPavlov  Nov 21, 2018
#2
0

4)

Factor the following:
4 (2 - a)^2 - 81

4 (2 - a)^2 - 81 = (2 (2 - a))^2 - 9^2:
(2 (2 - a))^2 - 9^2

Factor the difference of two squares. (2 (2 - a))^2 - 9^2 = (2 (2 - a) - 9) (2 (2 - a) + 9):
(2 (2 - a) - 9) (2 (2 - a) + 9)

2 (2 - a) = 4 - 2 a:
(2 (2 - a) - 9) (4 - 2 a + 9)

Grouping like terms, -2 a + 4 + 9 = (4 + 9) - 2 a:
(4 + 9) - 2 a (2 (2 - a) - 9)

4 + 9 = 13:
(2 (2 - a) - 9) (13 - 2 a)

2 (2 - a) = 4 - 2 a:
(4 - 2 a - 9) (13 - 2 a)

Grouping like terms, -2 a - 9 + 4 = -2 a + (4 - 9):
-2 a + (4 - 9) (13 - 2 a)

4 - 9 = -5:
(-5 - 2 a) (13 - 2 a)

Factor -1 out of -2 a - 5:

-(2a + 5) (13 - 2a)

Nov 21, 2018
#3
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1) Factor out (a      )(3a      )

See the multiples of -8 and have them add up to -10 when one of them is multiplied by 3.

You get $$(a-4)(3a+2)$$

2) The multiples of -3 are 1, -3 and -1, 3 and after trying out we get $$(15x^2+3)(2x-1)$$

3) We see that we can do a difference of squares. $$(4+5\sqrt x)(4-5\sqrt x)$$ .

4) This equals $$4(a^2-4a+4)-81=4a^2-16a+16-81=4a^2-16a-65$$

Factor $$4a^2-16a-65=(2a+5)(2a-13)$$ .

You are very welcome!

:P

(also thanks PartialMathematician for pointing out a mistake!)

Nov 21, 2018
edited by CoolStuffYT  Nov 21, 2018
#9
+701
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#3 could be factored into $$4^2 - 5^2(x)$$

PartialMathematician  Nov 21, 2018
#10
+701
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We could also substitute $$y^2$$ for $$x$$, so we have $$16-25y^2$$. Now, this looks looks like a difference of squares. We can factor it into $$(4+5y)\cdot(4-5y)$$. Substituting back $$x$$ for $$y^2$$, we have $$\boxed{(4+5\sqrt{x})\cdot(4-5\sqrt{x})}$$

Hope this helps,

- PartialMathematician

PartialMathematician  Nov 21, 2018
#12
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You are right, thanks.

CoolStuffYT  Nov 21, 2018
#6
+101322
+1

We can actually factor 3 as a difference of squares  if we note that

x  = √x * √x

16 - 25 x   =

(4 + 5√x ) ( 4 - 5√x )

Nov 21, 2018
#7
+101322
+1

Also

4) 4(2-a)^2 -81     is a difference of squares

[ 2 (2-a) + 9 ]  [ 2(2-a) - 9 ]   =

[4 - 2a + 9 ]  [ 4 - 2a - 9 ]    =

[13 - 2a ] [ -5 - 2a ] =       { factor a "-1" from the last term....move it to the front }

- [13 - 2a ] [ 5 + 2a ]

[ 2a - 13 ] [ 2a + 5 ]

Nov 21, 2018
#8
+18329
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Wooohoooo!  Nice, Chris !

ElectricPavlov  Nov 21, 2018
#11
+101322
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Thanks, EP.....sometimes....my brain actually works!!!

CPhill  Nov 21, 2018