Question

i would really like some help, factoring trinomials if you want an extra 50-100 points you could help with a previous question on my profile lol

Answer 1

`g) \left(5x-2\right)\left(3x+2\right)`

`h)2\left(8a-9\right)\left(a-2\right)`

`i)3\left(21n^2+42n+16\right)`

Explanation:

Starting with g)
`15^2+4x-4`

Break the expression into group

`=\left(15x^2-6x\right)+\left(10x-4\right)`

Now, Factor out
`3x` from
`15x^2-6x` which is now is
`3x(5x-2)`

Next, Facotr out
`2` from
`10x-4` which is now is
`2(5x-2)`

Thus,

`=3x\left(5x-2\right)+2\left(5x-2\right)`

Factor common term 5x -2

`\left(5x-2\right)\left(3x+2\right)`

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Next we have
`h)16a^2-50a+36`

Factor out common term thus we have
`2(8a^2-25a+18)`

Factor again:
`8a^2-25a+18` now turn into
`(8a-9)(a-2)`

`=2\left(8a-9\right)\left(a-2\right)`

-------------------------------------------------------------------------------------------------------------

Lastly we have
`i)63n^2+126n+48`

Rewrite the following:

63 as 3 * 21

126 as 2 * 42

48 as 3 * 16

`63n^2+126n+48`

Now cut out common term:

`=3\left(21n^2+42n+16\right)`

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~lenvy~

Answer 2

(g)
`(5x-2)(3x+2)`

(h)
`(16a-18)(a-2)`

(i)
`3(21n^2+42n+16)`

Explanation:

To factor a quadratic in the form
`ax^2+bx+c`

• Find 2 two numbers (
  `d` and
  `e`) that multiply to
  `ac` and sum to
  `b`
• Rewrite
  `b` as the sum of these 2 numbers:
  `d + e = b`
• Factorize the first two terms and the last two terms separately, then factor out the comment term.

Question (g)

`15x^2+4x-4`

`\implies ac=15 \cdot -4=-60`

`\implies d+e=4`

Factors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

Therefore, the two numbers (d and e) that multiply to -60 and sum to 4 are:

10 and -6

Rewrite
`4x` as
`+10x-6x`:

`\implies 15x^2+10x-6x-4`

Factories first two terms and last two terms separately:

`\implies 5x(3x+2)-2(3x+2)`

Factor out common term
`(3x+2)`:

`\implies (5x-2)(3x+2)`

Question (h)

`16a^2-50a+36`

`\implies ac=16 \cdot 36=576`

`\implies d+e=-50`

Factors of 576: 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 32, 36, 48, 64, 72, 96, 144, 192, 288, 576

Therefore, the two numbers that multiply to 576 and sum to -50 are:

-32 and -18

Rewrite
`-50a` as
`-32a-18a`:

`\implies 16a^2-32a-18a+36`

Factories first two terms and last two terms separately:

`\implies 16a(a-2)-18(a-2)`

Factor out common term
`(a-2)`:

`\implies (16a-18)(a-2)`

Question (i)

`63n^2+126n+48`

Factor out common term 3:

`\implies 3(21n^2+42n+16)`

This cannot be factored any further.