81x^3 – 192 First, factor out any common terms 3(27x^3 – 64) 3 ( (3x)^3 – (4)^3 ) 3(3x-4)(9x^2+12x+16)
Your goal is to factor pairs of terms that have a common monomial. Above is the pattern when grouping.
For factoring, usually try to factor out the most common term from the two highest variables. Then factor out the most common term from the other two variables. As with this example, you can now see that we have two pairs that are the same. We are able to factor like this because our x-3 is our like term. If we had x^2(a) – 16(a), we too would factor out our common term, the a. Don’t let the x-3 confuse you. It is just a common factor.
Exercise 1: x^3 + 2x^2 + 3x +6 x^2(x+2) + 3(x+2) (x^2+3)(x+2) Exercise 2: m^3 -2m^2+4m-8 m^2(m-2) + 4(m-2) (m^2+4)(m-2) To take this a step further, we can solve our polynomial by setting our factors equal to zero. m^2 +4 =0 or m-2=0 m^2 = -4 m=2 sqrt(m^2) = sqrt(-4) m = +2i, -2i
Try rewriting your first power so that it is being raised to the second power (so it’s squared). EX: x^4 = (x^2)^2 x^6 = (x^3)^2 x^8 = (x^4)^2 and so forth. Example: x^4 -81 = (x^2+9)(x^2-9) But, we’re not done yet. Our second factor is a difference of squares and can be factored further. =(x^2+9)(x+3)(x-3) Since we have no common terms or any factors that can be reduced, we are done!
Exercise 3: 16x^4 – 81 =(4x^2 +9)(4x^2-9) =(4x^2+9)(2x+3)(2x-3) Exercise 4: 6y^6 – 5y^3 -4 We can rewrite y^6 as (y^3)^2 to break up the term. (3y^2 -4)(2y^2+1)
Whenever you have a polynomial, the first thing you always want to check and see is what factors do you have in common. Your first step is to always factor out any common terms. This makes it a lot easier to factor and see how to factor your polynomial. In the above example, once we had our common term factored out, you can now see that we have a trinomial that needs factored in quadratic form.
V = l*w*h 24 = (4x)(x-1)(x-2) 24 = (4x^2 -4x)(x-2) 24 = (4x^3 – 8x^2 -4x^2 + 8x) 24 = 4x^3 -12x^2 + 8x 0 = 4x^3 – 12x^2 +8x – 24 Now, factor by grouping! 0 = 4x^2(x-3) + 8(x-3) 0 = (4x^2+8)(x-3) Once factored, set equal to zero and solve for x. This is similar to what we did in sections 4.3 and 4.4 when factoring quadratics. 4x^2 + 8 = 0 or x – 3 =0 4x^2 = -8 x = 3 x^2 = -2 sqrt(x^2) = sqrt(-2) x = i sqrt(2) Notice how our one solution is imaginary. What does this mean? Can we use this? No. Because it is imaginary, we can eliminate this solution. So, x = 3. To complete the problem, find the dimensions of our jewelry box. Length = 4x = 4(3) = 12 Width = (x-1) = (3-1) = 2 Height = (x-2) = (3-2) = 1 Dimensions: 12 by 2 by 1
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Bell Ringer
Factor the following:
81x 3 – 192
3(3x - 4)(9x 2 + 4x + 16)
(3x- 4)(9x 2 + 4x + 16)
3(3x - 4)(9x 2 + 12x + 16)
(9x - 4)(9x 2 + 4x + 16)
Students will be able to factor polynomial equations. Page 356 #18-29, 32-40 (even)
The dimensions of a jewelry box are: length 4x, width (x-1), and height (x-2). If the volume of the box is 24 cubic inches, find the dimensions of the box.
Hint: Remember V=lwh . Multiply this out, and then try factoring by grouping to solve.
State the new dimensions, and show all of your work.
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