How do you factor #x²-11x+30#?

3 Answers
Mar 18, 2018

Answer:

#(x-5)(x-6)#
Refer to explanation.

Explanation:

This is a quadratic equation in the form of #ax^2+bx+c#, so you simply factor b into two parts such that their sum is b and product is ac.

#x^2-11x+30#
#x^2-5x-6x+30#
#( -5xx-6=30 and -5-6=-11)#
#x(x-5)-6(x-5)#
#(x-5)(x-6)#

That's it.
Hope this helps :)

Mar 18, 2018

Answer:

#x^2-11x+30=color(blue)((x-5)(x-6))#

Explanation:

If we attempt to factor #x^2-11x+30#
into the product of two binomials: #(x+a)(x+b)#
since #(x+a)(x+b)=x^2+(a+b)x+ab#
we see that
#a# abd #b# must have the same sign (since the constant term is #color(red)+30#)
#ab# must be equal to #30#
#a+b# must add up to #-11# (which implies they both must be negative).

#{: ("negative factor pairs of "30":"," | ",-1xx-30," | ",-3xx-10," | ",color(blue)(-5xx-6)), ("sum of factor pairs:"" | ",,-31," | ",-13," | ",color(blue)(-11)) :}#

Mar 18, 2018

Answer:

#x=5# #,6#

Explanation:

On comparing this equation with the equation #ax^2+bx+c = 0#
#a = 1,b= -11,c= 30#
Now you have to split the middle term i.e. -11 in two parts such that the multiplication of both the parts is equal to #a*c# while the addition of both the parts is equal to #b# itself.
We can divide #-11# as #-6# and #-5#
So, the equation becomes
#(x^2 +(-6-5)x+30) = 0#
#(x^2-6x-5x+30) = 0#
Now taking out common terms
#(x[x-6]-5[x-6]) = 0#
#([x-5][x-6]) = 0#
#x=5 x=6#