Question

Please help!!!

Answer 1

Answer:

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

Answer:

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

Step-by-step 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

$\quad ac=3\times -7$

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. $\sf \quad group : \\ \quad \rm (3x - 7)(x + 1)$

$\text{And we are done!}$