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Start by finding the GCF of the expression. To do this, rewrite each term as the products of its factors. 4x^3 &= 2* 2 * x* x* x 6x^2 &= 2* 3 * x* x 12x &= 2* 2 * 3* x The GCF of the initial expression is 2 x.
Next, rewrite each term of the initial expression as the product of the GCF and another factor. 4x^3 &= 2 x* 2x^2 6x^2 &= 2 x* 3x 12x &= 2 x* 6 One way of finding the corresponding factors is dividing the original terms by the GCF. 4x^3/2 x &= 2x^2 [0.35em] 6x^2/2 x &= 3x [0.35em] 12x/2 x &= 6 Therefore, the initial expression can be written as follows. 4x^3 + 6x^2 - 12x ⇕ [0.25em] 2 x* 2x^2 + 2 x* 3x - 2 x* 6
Finally, using the Distributive Property, factor out the GCF. 4x^3 + 6x^2 - 12x ⇕ [0.25em] 2 x( 2x^2 + 3x - 6) After this, the expression between the parentheses has to be studied to continue with the factorization, if possible.
To factor an expression as a difference of two squares, the terms of the expression should be perfect squares. 9x^2-121 ⇓ ( 3x)^2-( 11)^2 As it can be seen, the terms of the expression are perfect squares.
One good way to recognize if a trinomial is a perfect square trinomial is to look at its first and last terms. If they are both perfect squares, there is a good chance that it is a perfect square trinomial. In the given expression, the first and last terms can be written as the squares of 4x and 11, respectively. 16x^2+88x+121 ⇓ ( 4x)^2+88x+( 11)^2 These perfect squares show that the expression could be a perfect square trinomial. However, this is not enough to decide yet.
The next step is to check whether the middle term is two times the square roots of the first and last terms. ( 4x)^2+88x+( 11)^2 ⇓ ( 4x)^2+2( 4x)( 11)+( 11)^2 It can be seen that the given expression satisfies this condition as well.
Since the expression satisfies both conditions, it is a perfect square trinomial. Therefore, it can be written as a square of a binomial where 4x and 11 are the first and second terms of the binomial, respectively. 16x^2+88x+121=( 4x+ 11)^2
To factor quadratic trinomials in the form x^2+bx+c, two numbers p and q with a sum of b and a product of c must be found so that the trinomial is written as the product of (x+p) and (x+q). x^2+bx+c = (x+p)(x+q)
As an example, the trinomial below will be factored. x^2+7x+12 These three steps can be followed to factor it.For some p and q, the aim is to write the given trinomial as follows. x^2+7x+12 = (x+p)(x+q) To do so, the signs of b and c will be used to determine the signs of p and q. x^2+ 7x+ 12 Here, b= 7 and c= 12, so both b and c are positive.
It is known that b=7, c= 12 and that p and q are positive integers. Therefore, two positive factors of 12 whose sum is 7 need to be found. The positive factor pairs of 12 will be listed and the pair with a sum of 7 identified.
Positive Factors of 12 | Sum |
---|---|
1 and 12 | 1+12=13 |
2 and 6 | 2+6=8 |
3 and 4 | 3+4=7 |
As seen, the factor pair of 3 and 4 meet these requirements, so the values of p and q are 3 and 4.
For the given trinomial, two integers with a sum of 7 and a product of 12 were found. x^2+7x+12 ⇒ lp=3 q=4 Therefore, the trinomial can be written as the product of the binomials x+3 and x+4. x^2+7x+12 = (x+3)(x+4)
seeits factors. Consider the following expression. 8x^2+ 26x+ 6 Here, a= 8, b= 26, and c= 6. There are six steps to factor this trinomial.
It is known that a= 4 and c= 3, so a c=12>0. Therefore, the factors must have the same sign. Also, b= 13. Since the sum of the factors is positive and they must have the same sign, both factors must be positive. All positive factor pairs of 12 can now be listed and their sums checked.
Factors of a c | Sum of Factors |
---|---|
1 and 12 | 1+12=13 ✓ |
2 and 6 | 2+6=8 * |
3 and 4 | 3+4=7 * |
In this case, the correct factor pair is 1 and 12. The following table sums up how to determine the signs of the factors based on the values of ac and b.
ac | b | Factors |
---|---|---|
Positive | Positive | Both positive |
Positive | Negative | Both negative |
Negative | Positive | One positive and one negative. The absolute value of the positive factor is greater. |
Negative | Negative | One positive and one negative. The absolute value of the negative factor is greater. |
Such analysis makes the list of possible factor pairs shorter.
Start by grouping the first two terms and the last two terms. y=2x^3-x^2+6x-3 ⇕ [0.25em] y=(2x^3-x^2) + (6x-3)
In the first pair, factor out the GCF, which is x^2. y = (2x^3-x^2) + (6x-3) ⇕ [0.25em] y = x^2(2x-1) + (6x-3) For the second pair, the GCF is 3. y = x^2(2x-1) + (6x-3) ⇕ [0.25em] y = x^2(2x-1) + 3(2x-1)
If the polynomial is factorable, the second step should lead to a sum of two terms with a common factor. y = x^2( 2x-1)+ 3( 2x-1) In this case, the GCF of the terms is ( 2x-1). Thus, factoring it out allows factoring the initial polynomial. y = x^2( 2x-1)+ 3( 2x-1) ⇕ [0.25em] y = ( 2x-1)( x^2 + 3)
There might be polynomials that require some rewrites before they can be factored by grouping. If the rewrite is not done, the method could not work even though the polynomial is factorable. For example, consider the following polynomial. y = x^3 - 3x^2 + 4x -2 Grouping the terms in pairs and factoring out their corresponding GCF will not help in factoring the polynomial. y = (x^3 - 3x^2) + (4x -2) ⇓ y = x^2(x-3) + 2(2x-1) As seen, the terms between parentheses are not equal and therefore, they cannot be factored out. However, rewriting -3x^2 as -x^2-2x^2 and 4x as 2x+2x could be useful. y = x^3 - 3x^2 + 4x - 2 ⇓ y = x^3 - x^2 - 2x^2_(-3x^2) + 2x + 2x_(4x) - 2 Now, there are six terms in the polynomial and they can be grouped in pairs. y = (x^3 - x^2) + (-2x^2 + 2x) + (2x - 2) Next, factor out the GCF of each pair. In this case, factor out x^2, -2x, and 2 from the first, second, and third pair respectively. y = x^2 (x-1) - 2x (x-1) + 2 (x - 1) Notice that (x-1) is a common factor for the three terms. Therefore, it can be factored out. y = (x-1)(x^2-2x+2) This way the initial polynomial was written as the product of two factors. To complete the factorization, the quadratic polynomial needs to be factored. Note that the rewrites made here are particular for the given polynomial.