8. Factoring Polynomials Completely
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To consider a polynomial factored completely, we need to be sure that it cannot be factored any further. Which methods can you use to do this?
See solution.
To consider a polynomial factored completely, we need to be sure that it cannot be factored any further. We will discuss the steps we can follow to achieve this, and then we will review the special product patterns and factoring methods we can use.
The GCF is the greatest monomial that we can factor out from every term in a polynomial expression. Let's consider the polynomial shown below. 9x^5+6x^3+12x^2 We can rewrite it to show explicitly that the greatest power of x in common is x^2, and the greatest number that divides exactly the three coefficients is 3. (9x^5)+(6x^3)+(12x^2) ⇕ ( 3* 3 * x^3 * x^2)+( 3 * 2 * x^2 * x) +( 4* 3* x^2) Therefore, the GCF is 3x^2. 9x^5+6x^3+12x^2 ⇔ 3x^2(3x^3+2x+4)
The product of a sum and difference of two terms gives as result the difference of the squares of the terms. (a+b)(a-b) =a^2-b^2 Therefore, if we have a polynomial that is the difference of two perfect squares, we can factor it as the product of a sum and difference of two terms. For example, consider the polynomial shown below. x^2-25 Note that x^2 and 25=5^2 are perfect squares whose square roots are x and 5. Therefore, this is a difference of squares and it can be factored as the product of a sum and difference of two terms. a^2-b^2 = (a+b)(a-b) ⇓ x^2-25 = (x+5)(x-5)
The process of squaring a binomial gives as result a perfect square trinomial. (a+b)^2 =a^2+2ab+ b^2 (a-b)^2 =a^2-2ab+ b^2 Therefore any perfect square trinomial can be factored as the square of a binomial. Note that a trinomial has to satisfy two conditions to be a perfect square trinomial.
Let's consider the example trinomial shown below. x^2+8x+16 Note that x^2 and 16=4^2 are perfect squares, whose square roots are x and 4. Hence, the first condition is satisfied. Furthermore the middle term, 8x, equals two times the product of the square roots of the others. 8x &= 2* x*4 8x &= 2* 4* x 8x &= 8x ✓ Therefore, the second condition is also satisfied, and we can factor the trinomial as the square of a binomial. x^2+8x+16 ⇔ (x+4)^2 Here the binomial is a sum of two terms. If the middle term of the trinomial was negative, we would have used a negative sign for the square of the binomial instead. x^2 - 8x+16 ⇔ (x - 4)^2
To factor a trinomial of the form x^2+bx+c, we need to look for a pair of factors of c that adds up to b. Let's consider the example trinomial x^2-x-2. x^2+ bx + c x^2+( - 1)x+ ( - 2) Since c= - 2, is negative, we need to look for factors of opposite signs. Furthermore, since b= - 1 is negative, we can know that the greatest factor should be the negative. Since the only factors of 2 are 2 and 1, there is only one possibility for the factors. The factored form is shown below. x^2-x-2 ⇔ (x-2)(x+1)
To factor a trinomial of the form ax^2+bx+c, we need to look for factors of a and c. Let's consider the example trinomial 6x^2-x-2. ax^2+ bx + c 6x^2+ (-1) x + ( - 2) Since c= - 2, is negative, the factors need to have different signs. We can use a table to organize and verify the possible factors we can use.
| Factors of 6 | Factors of - 2 | Possible Factorization | Middle Term |
|---|---|---|---|
| 6,1 | - 2,1 | (6x-2)(x+1) | - 2x+6x = 4x * |
| 6,1 | 2,- 1 | (6x+2)(x-1) | 2x -6x =- 4x * |
| 3,2 | - 2,1 | (3x-2)(2x+1) | - 4x +3x = - x âś“ |
| 3,2 | 2,- 1 | (3x+2)(2x-1) | 4x-3x = x * |
We have found the correct factored form for the trinomial given. 6x^2-x-2 ⇔ (3x-2)(2x+1)