factorise the following?

a. e^2-16= (e+4)(e-4) answer
b 2y^2-18=2(y^2-9=2(y+3)(y-3) answer
c. 3z^2+3z-18=3(z^2+z-6)
.......................=3(z+3)(z-2)answer

a^2 - b^2 = (a + b)(a - b)

a)
e^2 - 16
= e^2 - 4^2
= (e + 4)(e - 4)

b)
2y^2 - 18
= 2(y^2 - 9)
= 2(y^2 - 3^2)
= 2(y + 3)(y - 3)

c)
3z^2 + 3z - 18
= 3(z^2 + z - 6)
= 3(z^2 + 3z - 2z - 6)
= 3[(z^2 + 3z) - (2z + 6)]
= 3[z(z + 3) - 2(z + 3)]
= 3(z + 3)(z - 2)

To solve questions like the above, there are three steps;

1) place in the familiar format, ax^2 + bx + c, the ones above have already been placed into this format.

2) write down the brackets with the x:
(x ) (x )

3) Look for the factor pairs that multiply to give c and add to give b.

For example x^2-5x=-6

Add 6 to both sides to get x^2-5x+6=0

(x )(x )

Factors of 6; 1 6, 2 3. Remember these can be both + or -

1-6=-5 1.-6=-6
-2-3=-5 -2.-3=6 we are looking for +6

so (x-2)(x-3) is the solution.

a (e-4)(e+4)
b (2y-6)(y+3 )
c (3z+9 )(z-2 )

a. e^2-16
(e+4)(e-4)

b. 2y^2-18
2(y^2 - 9)
2(y+3)(y-3)

c. 3z^2+3z-18
3(z^2+z-6)
3(z-2)(z+3)

In b and c, you first have to factor out a common factor of 2 and 3 respectively and then you factor it.

Hope this helps.


(P.S.- Factoring just take a bit of practice of finding numbers that add up together and multiple together to work out for the equation. For instance in the last problem, -2 and 3 multiply together to get -6, and when you subtract them, you'll get +1. Which will give you the equation z^2+z-6. Also don't mind the variable, it can be e, y, or z as in your cases or x, b and c in other cases.)

a.e^2-16 = (e - 4)(e +4)

b.2y^2-18 = 2 (y^2 - 9) = 2(y+3)(y-3)

c. 3z^2+3z-18 = 3(z^2 + z - 6) = 3(z +3)(z - 2)