# factorise c^2-8c+15

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The task is to

### factorise c^2-8c+15

The "-8" coefficient of c suggest that the 2 factors that must be multiplied to obtain 15 must be negative (obeying the rule "-" multiplied by "-" will yield "+")

so,

c^2 - 8c + 15

= (c-5)(c-3)

**So the required answer, the factors of c^2 - 8c + 15 are (c-5) and (c-3)**

Click on the link provided at the bottom of the answer to appreciate the solution graphically.

Notice that the graph cuts the x-axis (or c-axis in our case) at x=3 and x=5. :)

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A note:

Many students will dive head-long into applying the standard formula {x=[-b+/- sqrt(b^2-4ac)]/2a} for the solution of quadratic equations. However, I would urge my students to tickle their minds (it's a good mind exercise!) to entertain the possibilities of having nice products of factors of "c/a" that will add up to "b/a". ie. if a is 1, we are trying to find the factors of "c" that will add up to "b"

c^2 - 8c + 15 = ( c -3)(x-5)

You can calculate the roots using the following formula:

x= [ -b +-sqrt(b^2 - 4ac)]/2a

a = 1 b= -8 c = 15

==? x1= [ 8 + sqrt(64-4*1*15) ]/ 2

= [ 8 + sqrt(4)]/2 = (8+2)/2 = 5

==> x1= 5

==> x2= ( 8-2)/2 = 6/2 = 3

==> x2= 3

We know that if x1 and x2 are roots for f(x) then ( x-x1) and (x-x2) are factors of f(x):

==> (x-5) and (x-3) are factors.

**==> ( c^2 - 8c + 15 = ( c-5)(c-3)**

To factorise c^2-8c+15.

c^2-8c needs 4^2 to make a perfect square as c^2-8c+4^2 = (c-8)^2.

Therefore we rewrite the given expression as below:

c^2-8c+15 = c^2-8c+16 -1

(c^2-8c+16) - 1^2 = (c-4)^2 - 1^2 . This expresssion is like a^2-b^2 which is equal to (a-b)(a+b).

Therefore (c-4)^2 -1^2 = (c-4 -1) (c-4 +1).

Therefore (c-4)^2 - 1= (c-5)(c-3).

Therefore the given expression c^2-8c+15 is equal to (c-5)(c-3) inthe factored form.