How do you factor #v^2+16v+63#?
1 Answer
Explanation:
Given:
#v^2+16v+63#
Here are a few methods...
Fishing for factors
Note that:
#(v+a)(v+b) = v^2+(a+b)v+abv#
So if we can find two numbers
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)#
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
#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.
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
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
Since the leading coefficient is
#v^2+16v+63 = (v+7)(v+9)#