#8x^2-14x+3#? I dont understand how to solve this using factoring grouping?

#8x^2-14x+3#

1 Answer
Mar 8, 2017

Answer:

Please see below.

Explanation:

It is important to know that when you try to factorize a quadratic polynomial say #ax^2+bx+c# by grouping,

  1. Remember you can do it, only if the determinant #b^2-4ac# is a complete square of a number . This is generally not a big deal as generally in problems where factoring by grouping is desired, this is ensured. However, if you are unable to do this you can check for this. Here in #8x^2-14x+3#, we have #a=8#, #b=-14# and #c=3#. Hence #b^2-4ac=(-14)^2-4xx8xx3=196-96=100=10^2# and you can factorize the given polynomial by grouping.

  2. Now you have to split #b# in two parts, whose sum is #b# and product is #axxc#. Here to make things easier look at the signs of #a# and #c# . If they are same, look at their sum equal to #b# and if signs are different, look at their difference equal to #b#.
    Here, signs of #a=8# and #b=3# are same, as they are bot positive (an example of different signs given below), hence find two numbers whose sum is #14# and product is #8xx3=24#.

  3. For making things easier, particularly when #axxc# is a large number, list factors as shown here #((1,24),(2,12),(3,8),(4,6))# and we end here as the factors are getting repeated there after. Here sum of #2# and #12# is equal to #14#, hence split accordingly.

Now #8x^2-14x+3#

= #8x^2-2x-12x+3#

= #2x(4x-1)-3(4x-1)#

= #(2x-3)(4x-1)#

Had the polynomial been #8x^2-5x-3#, as signs of #a# and #c# are opposite, and difference is desired to be #5#, we should have chosen #3# and#8# and factors would have been

#8x^2-5x-3#

= #8x^2-8x+3x-3#

= #8x(x-1)+3(x-1)#

= #(8x+3)(x-1)#