Question

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

`8x^2-14x+3`

Answer 1

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)`