The product of the polynomials (x + 2), (x – 2), (x 3– 2x 2+ 4x – 8) and (x 3+ 2x 2+ 4x + 8) is
(x 4– 16) 2
Calculating the Product of Polynomials
The question asks us to find the product of four given polynomials: \((x + 2)\), \((x - 2)\), \((x^3 - 2x^2 + 4x - 8)\) and \((x^3 + 2x^2 + 4x + 8)\).
Let's find the product by multiplying these polynomials step-by-step. We can group and factor some terms to simplify the calculation.
First, consider the product of the first two polynomials, \((x + 2)\) and \((x - 2)\). This is a classic difference of squares pattern:
\[(x + 2)(x - 2) = x^2 - 2^2 = x^2 - 4\]
Now let's look at the third polynomial: \((x^3 - 2x^2 + 4x - 8)\). We can try factoring this by grouping terms:
\[x^3 - 2x^2 + 4x - 8 = (x^3 - 2x^2) + (4x - 8)\]
Factor out common terms from each group:
\[= x^2(x - 2) + 4(x - 2)\]
Now, factor out the common binomial factor \((x - 2)\):
\[= (x^2 + 4)(x - 2)\]
So, \((x^3 - 2x^2 + 4x - 8) = (x^2 + 4)(x - 2)\).
Next, let's look at the fourth polynomial: \((x^3 + 2x^2 + 4x + 8)\). We can also factor this by grouping terms:
\[x^3 + 2x^2 + 4x + 8 = (x^3 + 2x^2) + (4x + 8)\]
Factor out common terms from each group:
\[= x^2(x + 2) + 4(x + 2)\]
Now, factor out the common binomial factor \((x + 2)\):
\[= (x^2 + 4)(x + 2)\]
So, \((x^3 + 2x^2 + 4x + 8) = (x^2 + 4)(x + 2)\).
Now we need to find the product of all four original polynomials. Substituting the factored forms, the product is:
\[(x + 2)(x - 2)(x^3 - 2x^2 + 4x - 8)(x^3 + 2x^2 + 4x + 8)\]
\[= (x + 2)(x - 2) \cdot [(x^2 + 4)(x - 2)] \cdot [(x^2 + 4)(x + 2)]\]
Let's rearrange the terms to group similar factors:
\[= [(x + 2)(x - 2)] \cdot [(x - 2)(x + 2)] \cdot [(x^2 + 4)(x^2 + 4)]\]
Using the results from our factorizations and the difference of squares identity:
\[= (x^2 - 4) \cdot (x^2 - 4) \cdot (x^2 + 4)^2\]
This simplifies to:
\[= (x^2 - 4)^2 (x^2 + 4)^2\]
We can use the property \((a \cdot b)^n = a^n \cdot b^n\) in reverse, i.e., \(a^n \cdot b^n = (a \cdot b)^n\):
\[= [(x^2 - 4)(x^2 + 4)]^2\]
Inside the square brackets, we again have a difference of squares pattern, where \(a = x^2\) and \(b = 4\):
\[(x^2 - 4)(x^2 + 4) = (x^2)^2 - 4^2 = x^4 - 16\]
Substitute this back into the expression:
\[= (x^4 - 16)^2\]
So, the product of the given polynomials is \((x^4 - 16)^2\).
Let's compare this result with the given options:
- \(x^8 - 256\)
- \((x^4 - 16)^2\)
- \((x^4 + 16)^2\)
- \((x^2 - 4)^2\)
Our calculated product matches option 2.
The final answer is \((x^4 - 16)^2\).
Revision Table: Key Concepts
| Concept | Description | Formula/Example |
|---|---|---|
| Difference of Squares | A binomial of the form \(a^2 - b^2\) can be factored into the product of \((a-b)\) and \((a+b)\). | \(a^2 - b^2 = (a - b)(a + b)\) Example: \(x^2 - 4 = (x - 2)(x + 2)\) |
| Factoring by Grouping | A technique used to factor polynomials with four or more terms by grouping terms and factoring common factors from each group. | \(ax + ay + bx + by = a(x + y) + b(x + y) = (a + b)(x + y)\) Example: \(x^3 - 2x^2 + 4x - 8 = x^2(x - 2) + 4(x - 2) = (x^2 + 4)(x - 2)\) |
| Exponent Rule: \((ab)^n\) | The power of a product is the product of the powers. | \((a \cdot b)^n = a^n \cdot b^n\) |
Additional Information: Polynomial Multiplication and Factoring
Multiplying polynomials involves distributing each term of one polynomial to every term of the other polynomial and then combining like terms. Factoring polynomials is the reverse process of multiplication; it involves breaking down a polynomial into a product of simpler polynomials.
- When multiplying binomials like \((a+b)(c+d)\), we use the FOIL method (First, Outer, Inner, Last) or simply distribute: \(ac + ad + bc + bd\).
- Special products like the difference of squares \((a-b)(a+b) = a^2-b^2\) or the square of a binomial \((a+b)^2 = a^2+2ab+b^2\) are useful shortcuts.
- Factoring techniques include finding the greatest common factor (GCF), factoring by grouping (as used in the solution), factoring trinomials, and using special factoring patterns like difference of squares, sum/difference of cubes, etc.
- In this problem, recognizing the factorable structure of the cubic polynomials \((x^3 - 2x^2 + 4x - 8)\) and \((x^3 + 2x^2 + 4x + 8)\) using factoring by grouping was crucial. Each factored into a product involving \((x^2 + 4)\) and either \((x-2)\) or \((x+2)\).
- Combining the factors strategically allowed us to repeatedly apply the difference of squares identity, simplifying the calculation significantly.
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