Okay, we have: A = (H/2)(b1+ b2) Multiply both sides by 2/H: 2A/H = b1 + b2 Subtract b2 from both sides: 2A/H - b2 = b1

@MarkFL Word problem woes The length of the rectangle is 9ft less than twice the width of the rectangle. The peimeter of the rectangle is 54 ft. Find the width and the length

@MarkFL Hey Mark quick question. In 3x^3-5x^2=0 I know we factor out the x^2 so it's x^2(3x-5)=0 But what do we do with the x^2? The rest is easy, just when I checked it magically disappears haha.

You want to set each factor equal to zero and solve for x, to get all of the solutions...so solve these equations: x^2 = 0 3x - 5 = 0 What do you get?

Yes, the "zero factor property" states that if the product of two or more factors is zero, then at least one of the factors must be zero. So, if we can find the values of x that make each factor zero, we find all of the values of x that make the original product equal to zero.

Interesting, I guess my brain is trying to get over the "It's attached to the bracket" so treat it as part of the factor. Then I checked the next question as 2x^3-8x, pulled out 2x(x^2-4)=0 , 2x=0, x=0/2=0, x=0, (x+2)(x-2)=0 x=-2,2. Seems like the right steps?

You rocked it, bro! Just a note: when you have gotten to: 2x(x^2 - 4)=0 Continue factoring: 2x(x+2)(x - 2) = 0 Now, at this point you can consider 2 and x to be separate factors...and setting 2 = 0 isn't going to yield anything (because it can never be true), so you can just set: x = 0 x + 2 = 0 x - 2 = 0 Solving these gives you, as you also found: x = -2, 0, 2

Sometimes you may get factors that can't be further factored into linear factors (factors of the form ax + b). So, you then use other methods for determining the roots of those factors, such as the quadratic formula if the factor is quadratic.

There are also cubic and quartic formulas, but they are extremely cumbersome. For those that don't factor or have rational roots, I will use the Newton-Raphson method from differential calculus to determine approximations for the roots.