How do you factor #v^2+16v+63#?

1 Answer
Jun 12, 2017

#v^2+16v+63 = (v+7)(v+9)#

Explanation:

Given:

#v^2+16v+63#

Here are a few methods...

#color(white)()#
Fishing for factors

Note that:

#(v+a)(v+b) = v^2+(a+b)v+abv#

So if we can find two numbers #a, b# with sum #16# and product #63# then we can factor the given quadratic.

We might try:

#8, 8 -> 8*8 = color(red)(cancel(color(black)(64)))#

#7, 9 -> 7*9 = 63#

Hence:

#v^2+16v+63 = (v+7)(v+9)#

#color(white)()#
Completing the square

The difference of squares identity can be written:

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

Complete the square, then use this with #a=(v+8)# and #b=1# as follows:

#v^2+16v+63 = v^2+16v+64-1" "# (where #64 = (16"/"2)^2#)

#color(white)(v^2+16v+63) = (v+8)^2-1^2#

#color(white)(v^2+16v+63) = ((v+8)-1)((v+8)+1)#

#color(white)(v^2+16v+63) = (v+7)(v+9)#

One advantage of this method is that it will always work, regardless of whether the coefficients of the factors are integers, rational numbers, irrational numbers or complex numbers.

#color(white)()#
Quadratic formula

This method is overkill for this particular problem, but is as powerful as completing the square.

Note that:

#v^2+16v+63#

is of the form:

#av^2+bv+c#

with #a=1#, #b=16# and #c=63#.

Its zeros are given by the quadratic formula:

#v = (-b+-sqrt(b^2-4ac))/(2a)#

#color(white)(v) = (-16+-sqrt(16^2-4(1)(63)))/(2*1)#

#color(white)(v) = (-16+-sqrt(256-252))/2#

#color(white)(v) = (-16+-sqrt(4))/2#

#color(white)(v) = (-16+-2)/2#

#color(white)(v) = -8+-1#

That is:

#v = -7" "# or #" "v = -9#

So both #(v+7)# and #(v+9)# are factors of #v^2+16+63#

Since the leading coefficient is #1#, we can deduce that:

#v^2+16v+63 = (v+7)(v+9)#