Question

Please help!!!

Factor the following

`3x^2-4x-7`
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Answer 1

3x²-4x-7=0

doing middle term factorization

3x²-7x+3x-7=0

x(3x-7)+1(3x-7)=0

(3x-7)(x+1)=0

Answer 2

`\huge \colorbox{red}{ \rm (3x - 7)(x + 1)}`

Explanation:

to understand this

you need to know about:

• factoring
• PEMDAS

tips and formulas:

• quadratic expression:ax²+bx+c

factoring quadratic:

1. Find two numbers that multiply to give ac, and add to give b
2. rewrite the middle with those numbers
3. factor out common terms
4. group

let's solve:

a is 3 and b is -7

`\quad ac=3* -7`

therefore

`\quad ac=-21`

now we need to find two numbers that give -21

to do so we need to find the factors of 21

which are

`\quad 1,3,7`

likewise,

`\quad -1,-3,-7`

in this case we can take any two numbers from negative and positive factors that give us -21 and -4

3 and -7 are the two numbers that multiply to give -21 (3×-7=-21) and add to give -4. (3+(-7)=-4)

now let's factor:

1. 
   `\sf rewrite \: - 4x \: as \: 3x - 7x : \\ \rm {3 x}^(2) + 3x - 7x - 7`
2. 
   `\sf factor \: out \: 3x : \\ \rm 3x(x + 1) - 7x - 7`
3. 
   `\sf factor \: out \: - 7 : \\ \rm 3x(x + 1) - 7(x + 1)`
4. 
   `\sf \quad group : \\ \quad \rm (3x - 7)(x + 1)`

`\text{And we are done!}`