Question

Factorise `3x^2+2x-3` ?

Answer 1

`y = 3(x + (1 - sqrt10)/3)(x + (1 + sqrt10)/3)`

Explanation:

`y = 3x^2 + 2x - 3`
`D = d^2 = b^2 - 4ac = 4 + 36 = 40` --> `d = +- 2sqrt10`
There are 2 real roots:
`x1 = -b/(2a) +- d/(2a) = - 2/6 +- (2sqrt10)/6 = - (1 - sqrt10)/3`,
`x2 = - (1 + sqrt10)/3`
Factored form:
`y = a(x - x1)(x - x2)`
`y = 3(x + (1 - sqrt10)/3)(x + (1 + sqrt10)/3)`

Answer 2

`3(x+(1-sqrt(10))/3) (x+(1+sqrt(10))/3)`

Solution in great detail.

Explanation:

Set `y=0=3x^2+2x-3" ".........................Equation(1)`

Compare to `y=ax^2+bx+c`

`x=(-b+-sqrt(b^2-4ac))/(2a)`

Set `a=3; b=2; c=-3`

`=> x=(-2+-sqrt(2^2-4(3)(-3)))/(2(3))`

`x=-1/3+-sqrt(40)/6`

`x=-1/3+-(cancel(2)^1sqrt(10))/cancel(6)^3`

`x=-1/3+-sqrt(10)/3`

`x=(-1+sqrt(10))/3 and x=(-1-sqrt(10))/3`

`(x+(1-sqrt(10))/3) (x+(1+sqrt(10))/3) " "..............Equation(2)`
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`color(blue)("Testing "Equation(2))`

`x (x+(1+sqrt(10))/3)color(white)("d") +color(white)("d")(1-sqrt(10))/3 (xcolor(white)("d")+color(white)("ddd")(1+sqrt(10))/3)`

`x^2+x/3+cancel((xsqrt(10))/3) color(white)("ddd")+color(white)("ddd")x/3-cancel((xsqrt(10))/3)+color(white)("d")(1^2-(sqrt(10))^2)/9`

`x^2+(2x)/3+1/9-10/9`

`x^2+(2x)/3-1`

So `Equation(2)` is not the same value as `Equation(1)` To make it do so multiply everything by 3

`y=3(x^2+(2x)/3-1) = 3x^2+2x-3 color(red)(larr" As required"`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Thus the answer is:

`3(x+(1-sqrt(10))/3) (x+(1+sqrt(10))/3)" ".........Equation(2_a)`