2016年2月28日 · There are two terms in this expression: 12x 2 and (-16x). First find the GCF to factor it out [note: that is, if the GCM is a factor of both terms, then it is a factor of the entire expression]. The terms are:

2015年7月13日 · Given #x^2-12x-2 = 0# Completing the square is simpler if we move the constant to the right side: #color(white)("XXXX")# #x^2-12x = 2# If #x^2# and #-12x# are the first two terms of a squared binomial expansion: #color(white)("XXXX")# #(x-a)^2 = (x^2-2ax+a^2)# then #color(white)("XXXX")# #a=6# and we need to add #a^2 = 36# to both sides to ...

2016年3月1日 · Yes these do. Rewriting equation (2) with split terms we obtain #(3x^2-12x+2x-8)# Now pair first two, and pair last two. Find out the common factor in each pair. Must take care of #-ve# sign in front of any term while pairing and inserting brackets. #(3x^2-12x)+(2x-8)#

2018年7月7日 · #sqrt12*sqrt(x^2)# The square root of something squared will just simplify to that something, and as for #sqrt12#, we can try to factor out a perfect square. We can rewrite #sqrt12# as. #sqrt4*sqrt3#, which simplifies to #2sqrt3#. Thus, we are left with #2sqrt3x#, or #2xsqrt3# Hope this helps!

2014年5月12日 · 12x 4 +10x 3-12x 2 First, let see if the terms all have a common factor. All three have 2x 2 as a factor.

12x 2 + 11x + 2 = 0 (12)(2) = 24 Find 2 numbers whose product is 24 and whose sum is 11 (8 and 3 work)

2019年3月19日 · If you want to factor a(x) = 8x 3 - 12x 2 + 18x -27, there are a few different methods you can try. Since there are four terms in the equation, the first option to try is factoring by grouping. a(x) = 8x 3 - 12x 2 + 18x -27. First, we want to make two groups of two terms each. a(x) = (8x 3 - 12x 2) + (18x -27 )

2018年6月1日 · x=2/3" or "x=1/4 >"rearrange into "color(blue)"standard form";ax^2+bx+c=0 "subtract "11x" from both sides" 12x^2-11x+2=0 "using the a-c method to factor the quadratic" "the factors of the product "12xx2=24 "which sum to - 11 are - 8 and - 3" "split the middle term using these factors" 12x^2-8x-3x+2=0larrcolor(blue)"factor by grouping" color(red)(4x)(3x-2)color(red)(-1)(3x-2)=0 "take out the ...

2021年5月4日 · This is an antiderivative initial value problem. Start by finding f '(x) and use f '(0) = 14 to find +C. Then use the same process to find f(x) using f(0)=3 to find the second C.

2017年2月22日 · a^2-12a+32=0 (x^2+4)^2+32=12x^2+48 is equivalent to (factorizing RHS) (x^2+4)^2+32=12(x^2+4) But as a=x^2+4, this can be written as a^2+32=12xxa or a^2+32=12a or a^2-12a+32=0