Factorise each of the following algebraic expressions completely,
(a) 2k-6k2 + 4k3
(b) 2ax - 4ay+ 3bx - 6by

see explanation

Step-by-step explanation:

(a)

Given

2k - 6k² + 4k³ ← factor out 2k from each term

= 2k(1 - 3k + 2k²)

To factor the quadratic

Consider the factors of the product of the constant term ( 1) and the coefficient of the k² term (+ 2) which sum to give the coefficient of the k- term (- 3)

The factors are - 1 and - 2

Use these factors to split the k- term

1 - k - 2k + 2k² ( factor the first/second and third/fourth terms )

1(1 - k) - 2k(1 - k) ← factor out (1 - k) from each term

= (1 - k)(1 - 2k)

1 - 3k + 2k² = (1 - k)(1 - 2k) and

2k - 6k² + 4k³ = 2k(1 - k)(1 - 2k)

(b)

Given

2ax - 4ay + 3bx - 6by ( factor the first/second and third/fourth terms )

= 2a(x - 2y) + 3b(x - 2y) ← factor out (x - 2y) from each term

= (x - 2y)(2a + 3b)

the f

last graph i believe

step-by-step explanation:

worked it out i believe

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18