Question

Find two values of "b" so that x^2 + b x + 6 can be factored into binomial factors.​

Answer 1

Given:

The expression is

$x^2+bx+6$

To find:

The values of b so that the given expression can be factored into binomials factors.

Solution:

An expression is $ax^2+bx+c$ factorable if b is the sum of possible factors of ac.

We have,

$x^2+bx+6$

Here, $a=1,b=b,c=6$.

$ac=(1)(6)$

$ac=6$

Some, factor forms of 6 are (1×6) and (2×3).

$1+6=7$

$2+3=5$

For b=7,

$x^2+7x+6=x^2+x+6x+6$

$x^2+7x+6=x(x+1)+6(x+1)$

$x^2+7x+6=(x+1)(x+6)$

For b=5,

$x^2+5x+6=x^2+2x+3x+6$

$x^2+5x+6=x(x+2)+3(x+2)$

$x^2+5x+6=(x+2)(x+3)$

Therefore, the two possible values of b are 7 and 5.