Ok the factors for 14 are 1&14 and 2&7.

Now look at the 5x and try to figure out which of the factors of 14 can be added or subtracted to equal 5. Cool, 2&7 are exactly what we are looking for because 7-2=5.

Don’t forget the x^2.

Now we have all the ingredients:

x times x = x^2

2 times 7 = 14

We have to set these up now ====> (x _ 7) times (x _ 2).

Notice that I haven’t put in the signs yet because I need to figure out how to make the +5x. So now the possibilities are:

(x+7)(x+2) or (x+7)(x-2) or (x-7)(x+2) or (x-7)(x-2)

Which one works? Hopefully you already know how to multiply these. Obviously (x+7)(x-2) is th one that works because:

x*x=x^2 and x*(-2)= -2x and (+7)*x= +7x and (+7)*(-2)= -14

now we add these guys together

x^2 + (-2x) + (+7x) + (-14) = x^2 + 5x – 14

Finally you have all your factors: (x+7) and (x-2)

take these separately and make them equal to zero.

x+7=0 and x-2=0 Now solve for x.

x=(-7) and x=(+2)

Sorry for the long answer but I hope it helps.

( ) ( ). Now, since you see that the x^2 has a coeficiant of one, the the x in both factors have to have coefficiants that when multiplyed equal 1. So the most likely look should now be (x ) (x )

Now look at the last number, the constant. This constant is found when you multply the last number in the two factors together. So we see this number is -14. So the multiples can be +/- 1&14 or +/- 2&7. Let’s leave those muliples for now and look at the 5 in the 5x. This number is formed when the x multiplies with the two multiples of the constant. So what two number equal -14 when multpliwd and equal 5 when added? -2 and 7. So let’s fill that in your final thing should look like this (x-2)(x+7)

mutiples of 14

7*2,2*7

step2

taking multiples as 7&-2

step3

now the quadratic equation can be written an

x^2+7x-2x-14=0

x(x+7)-2(x+7)=0

(x-2)(x+7)=0

now equating (x-2)=0& (x+7)=0=>x=2,-7

because x times x is x squared whic is the first term. then u get -2x and 7x so when u add that u get the middle term 5x. then 7 times -2 is -14 which is the last term.

then

x+7=0

x=0-7

x=-7

and

x-2=0

x=0+2

x=2

If not then you need to find 2 numbers which multiply to give -14 and add to give +5.

So we have

(x + 7)(x – 2)=0

x+7=0

x=-7

x-2=0

x=2

{5X=7X-2X and 7*(-2)=(-14)}

X^2+7X-2X-14=0

(X^2-2X)+(7X-14)

X(X-2)+7(X-2)=0

(X-2)(X+7)=0

X=2,-7

x2=53

i can t say how i get the answer=))))