Question

Factor the following expressions using your method of choice. After factoring each expression completely, check your

Answer 1

Answer:

1. $2x^{2}-x-10 =(2x-5)(x+2)$

2. $6x^{2}+7x-20 = (3x-4)(2x+5)$

3. $4x^{2}-4x+1 = (2x-1)(2x-1) = (2x-1)^{2}$

4. $A=\frac{1}{2}(2x-3)(x+3)=\frac{1}{2}(b*h)$

b = 2x-3 and h = x+3.

Step-by-step explanation:

1. First, we need to find two numbers that multiply together to get -10 and 2x².

For example we chose this combination:

2x² = 2x*x let's define a=2x and b=x

-10 = (-5)*2 let's define c=-5 and d=2

Now we must verify that a*d + b*c = -x (-x is the middle term of the polynomial)

2*2x + (-5)*x = 4x - 5x = -x, so these values are correct.

Then we can rewrite the polynomial as:

$2x^{2}-x-10 = (a+c)(b+d)=(2x-5)(x+2)$

2. We can use the same method.

6x² = 3x*2x let's define a=3x and b=2x

-20 = (-4)*5 let's define c=-4 and d=5

Now we must verify that a*d + b*c = 7x

3x*5 + 2x*(-4) = 15x - 8x = 7x, it is correct!

Then we can rewrite the polynomial as:

$6x^{2}+7x-20 = (3x-4)(2x+5)$

3. Before factoring it, we can multiply by -1 each term, we will have $4x^{2}-4x+1$

4x² = 2x*2x let's define a = 2x and b = 2x

1 = (-1)*(-1) let's define c = -1 and d = -1

Now we must verify that a*d + b*c = -4x

2x*(-1) + 2x*(-1) = -4x, it is correct!

Then we can rewrite the polynomial as:

$4x^{2}-4x+1 = (2x-1)(2x-1) = (2x-1)^{2}$

4. $A=x^{2}+\frac{3}{2}x-\frac{9}{2}$

If we factor out 1/2 of each term, we will have:

$A=\frac{1}{2}(2x^{2}+3x-9)$

And using the above method we can factorize

$2x^{2}+3x-9=(2x-3)(x+3)$

So, the area will be:

$A=\frac{1}{2}(2x-3)(x+3)$

and the area of the triangle is base times height over 2 ((b*h)/2)

$A=\frac{1}{2}(2x-3)(x+3)=\frac{1}{2}(b*h)$

Therefore, b = 2x-3 and h = x+3.

I hope it helps you!