Show me how to factorise k^2-5k+4?

k^2-5k+4 = k^2-4k-k+4

= k(k-4)-1(k-4)

= (k-1)(k-4)

Factor of k^2-5k+4 is (k-1)(k-4)

(k - 1)*(k - 4)

Numbers which multiply to give 4... 2,2 or 1,4
Both positive or both negative, as it's +4, not -4.
They must add to give -5,
A little trial and error leads to -1 and -4.
Working it backwards- have you done FOIL?
(x + a)*(x + b) = x^2 + a*x + b*x + a*b

the short version answer is : (k-4)(k-1)

In detail you most find two numbers that their some is equal to coefficient of k (i.e. -5) and their product is equal to to the constant (i.e. +4). Obviously the answer is -1 and -4 . And ...

The ac method of factories takes the form of given ax² +bx + c, find the factors of the product ac that sum to equal b, then use those factor to factor via grouping.

to put in algebraic form solve this system of equations ac = nm, n + m = b

so in k² − 5k + 4
a = 1, b = -5, c = 4
=> ac = 4

now -1 * -4 = 4
-1 + -4 = -5

=> k² − k − 4k + 4
=> k(k − 1) − 4(k − 1)
now (k − 1) is a common factor
=> (k − 1)(k − 4)

now for speed when a = 1 then this will always hold (x + n)(x + m) = x² + (n + m)x + nm

k^2 - 5k + 4

COEFFICIENT OF k^2 = +1

( +1) ( +4) = +4

FIND FACTORS OF +4 WHICH ADD UP TO - 5 ( COEFFICIENT OF k )

( -1) ( -4) = +4

k^2 -k -4k +4

k ( k-1) -4 ( k-1)

( k-4) ( k-1) ANSWER

first the k^2 can only be factored as k * k
so we have (k ?)(k ?)
the +4 tells us that we need 2 factors of 4 , that SUM to the 5 of the 5k
factors of 4 are 2,2 and 4 , 1
so the factors we are looking for are 4 and 1
the + in the +4 tells us the factors are of the same sign
the - from the -5 tells us that they are both negative

so we have (k-4)(k-1)

k² - 5k + 4 = (k - 4)(k - 1)

The simple way is to find two numbers that multiply to give 4 and combine with k to give -5k
(-1) (-4) = 4
(-1) + (-4) = (-5)

Here is a step by step solution:
http://www.symbolab.com/solver/basic-ope...

Hope it helps