How To Factorise 2 X-3 =0 Into Two Brackets

How do you factorise 3x^2 + 5x+2?

There is a method so you can obtain (x-1) (3x+2)3x^2 +5x +2Take the first coefficient a, and the last coefficient c,a=3 and c=2Times them together =6Now find which factors of 6 add up to coefficient b. b=52x3=6 and 2+3=5Now split up the equation:3x^2 +3x +2x +2 =0Factorise the first two terms, 3x^2 +3x3x(x+1)Then, just write down the bracket again next to it.3x (x+1) ___ (x+1)Fill in the gap, what times x+1 is 2x+2?3x(x+1) +2(x+1)Now finally, write the bracket (x+1) and put whatever is left in another bracket.(X+1) (3x+2)This method works for most quadratics with a coefficient a which is more than 1.

How do you factor (x^3 + 8)?

Polynomials do not have a guaranteed “easy” way to factor them. Therefore you must start with inspection meaning you look for anything that might be a clue as to how to proceed. This is where your mathematical skill is tested.Notice that [math]2^3 = 8[/math]. This is the clue. If you pick [math]x = -2[/math] then the equation equals 0. Therefore -2 is a root of the equation.Now your mathematical knowledge is tested. The fundamental theorem of algebra tells you that there is a simple relationship between the factors of a polynomial and its roots. In particular, if [math]a[/math] is a root of an equation then [math](x - a)[/math] is a factor. So we can remove one factor immediately.[math]a = -2[/math] so the factor is [math](x + 2)[/math]. We can write the whole equation with the remaining polynomial, or remainder: [math](x + 2)R(x) = x^3 + 8[/math]. Now we can actually use polynomial long division to find [math]R(x)[/math]. In polynomial long division you match up powers of x starting with the largest power and tracking the remainder just as if you were doing long division on a number.[math]\frac{x^3 + 8}{x+2}[/math]Apparently the latex command for long division isn’t included in quora so I can’t reproduce the long division. But the steps go like this:1. Quotient [math]x^2[/math]. Subtract [math](x^3 +2x^2)[/math] from the dividend. New dividend is [math]-2x^2 + 8[/math].2. Quotient [math]-2x[/math]. Subtract [math](-2x^2 -4x)[/math] from the dividend. New dividend is [math]4x + 8[/math].3. Quotient [math]4[/math]. Subtract [math](4x + 8)[/math] from the dividend. Remainder is 0. (the remainder must be zero since we know [math](x + 2)[/math] is a factor.The quotient now shows: [math]x^2 - 2x + 4[/math] which is the other factor R(x) allowing us to write:[math]x^3 +8 = (x+2)(x^2 - 2x + 4)[/math]But how do we know that [math]x^2 - 2x + 4[/math] cannot be factored? Well by inspection we see that this is an increasing function (never decreases) so it can only cross zero once and therefore has no other real factors. If we wanted to find the complex factors we could simply complete the square:[math]0 = x^2 - 2x + 4 = x^2 - 2x + 1 + 3 = (x - 1)^2 + 3[/math][math]-3 = (x - 1)^2[/math][math]x = 1 \pm i \sqrt{3}[/math]so we can write [math]x^3 +8 = (x+2)(x- 1 - i \sqrt{3})(x- 1 + i \sqrt{3})[/math]