Question

What is the value of `k` in the equation `6x^2-11x-10=(3x+2)(2x-k)`?

Answer 1

`k=5`

Explanation:

Expand the factors on the right side, using for example, the FOIL method.

`(3x+2)(2x-k)=6x^2-3kx+4x-2k`

`=6x^2+x(-3k+4)-2k`

Compare this to the left side. For the 2 sides to be equal, then

`-2k=-10rArrk=5`

Answer 2

`k=5`

Explanation:

The value of `color(violet)k` is determined by expanding the factors and then comparing the coefficients of the similar monomials(i.e monomials having same unknowns)

The expansion is determined by applying the distributive property
`color(red)((a+b)(c+d)=ac+ad+bc+bd)`

`6x^2-11x-10=(3x+2)(2x-k)`
`rArr6x^2-11x-10=(3x*2x+3x*(-k)+2*2x+2*(-k))`
`rArr6x^2-11x-10=6x^2-3xk+4x-2k`
`rArr6color(blue)(x^2)-11color(orange)x-10=6color(blue)(x^2)+(-3k+4)color(orange)x-2k`

Then,
`-3k+4=-11` EQ1
`-2k=-10rArrcolor(violet)(k=(-10)/(-2)=5)`

Checking the value of `color(violet)k` is determined by substituting its value in EQ1

`-3k+4=?-11`
`-3(5)+4=?-11`
`-15+4=?-11` TRUE

Therefore, `color(violet)(k=5)`