How was this equation factored?

Question

I'm really confused as to how $\frac 21(x^2-3x+\frac 94)$ was factored into $\frac 21(x-\frac 32)^2$.

If possible, explain it to me like I'm five.

I also don't quite get how $(x^2) = (x)^2$ is the same thing. What if $x$ is a negative?

Answer 1

Your two expressions are: $2(x^2 - 3x + 9/4)$ and $2(x - 3/2)^2$. Since each expression is written as the product of $2$ and a parenthetical expression, it will suffice to show the parenthetical expressions are equal. In other words, let us show that: $x^2 - 3x + 9/4 = (x - 3/2)^2$.

To do so, we will expand $(x - 3/2)^2$ using the distributive property to see if we get $x^2 - 3x + 9/4$.

Here we go:

$$(x - 3/2)(x - 3/2) = (x - 3/2)x + (x - 3/2)(-3/2) = (x^2 - 3x/2) + (-3x/2 + 9/4)$$

The final two expressions can be combined because there are like terms:

$$x^2 - 3x/2 -3x/2 + 9/4 = x^2 - 6x/2 + 9/4 = x^2 - 3x + 9/4$$

just as we hoped, wished, and desired. This establishes that the two parenthetical expressions are, indeed, equal; so, the two expressions multiplied by $2$ are equal, as well.

Answer 2

$$x^2-3x+\frac94=x^2+2\cdot\Big(-\frac32\Big)\cdot x+\Big(-\frac32\Big)^2=\Big(x-\frac32\Big)^2$$

Answer 3

It is important to realize that $(a + b)^2$ DOES !!!!!!NOT!!!!!! equal $a^2 + b^2$ so you wouldn't expect $(x - \frac 32)^2$ to equal $x^2 + \frac 94$. Something extra must APPEAR.

$(a + b)^2 = (a+b)(a+b) = a(a+b) + b(a + b) = (a*a + a*b)+(b*a + b*b)= a*a + (a*b + b*a) + b*b = a^2 + 2ab +b^2$.

So if you do $(a+b)^2 \implies a^2 + b^2 + SOMETHING $ the $SOMeTHING = 2ab$ must appear (out of nowhere).

So if you go the other way $a^2 + 2ab +b^2= (a+b)^2$ the $2ab$ must "disappear.

....

Okay?

So. Remember this $(a+b)^2 = a^2 + 2ab + b^2$.

Do $(a+b)^2 = (a+b)(a+b) = a(a+b) + b(a+b) = a^2 +(ab + ab) + b^2 = a^2 + 2ab + b^2$ enough times untill you absolutely convince yourself it is true.

Okay.

So How do we do something like $(x^2 - 3x + \frac 9/2) = (x \pm something) \pm something else?$.

Well we want $x^2 - 3x + \frac 94 = (x + a)^2= x^2 +2ax + a^2$

And we know $(x \pm a)^2 = x^2 \pm 2ax + a^2$.

So we need to set $x^2 = x^2$.

$-3x = 2ax$

$a^2= \frac 94 $

$-3x = 2ax \implies 2a = -3 \implies a=-\frac 32$.

And we also have $a^2 = \frac 94$ so $a =\pm 32$.

So we have $(x - \frac 32)^2 = x^2 -2*\frac 32 x + (\frac 32)^2$

$= x^2 - 3x + \frac 94$.