Common factor: 2nd Level
In a polynomial, the leading term is the term with the highest exponent. Normally, it is the first term on the left. In this polynomial,
3x² − 6x + 9,
3x² is the leading term.
Now we like the leading term to be positive. (We will see that when we factor trinomials.) Therefore, if we have the following,
−3x² + 6x − 9,
we can make the leading term positive by writing
−3x² + 6x − 9 = −3(x² − 2x + 3).
We can remove a negative factor.
Problem 10. Make the leading term positive.
| a) | −3x − 6 = −3(x + 2) | b) | −5x² + 5x = −5x(x − 1) | |
| c) | −x5 − x3 = −x3(x² + 1) | d) | −2x² + 6x − 2 = −2(x² − 3x + 1) | |
e) −32x3 − 12x² + 8x = −4x(8x² + 3x − 2)
f) −9x5 + 30x4 − 3x3 = −3x3(3x² − 10x + 1)
Problem 11. In each sum, remove the factor xn by displaying it on the left.
For example, xn + xn + 2 = xn(1 + x²).
| a) | xn + 2 + xn + 3 = xn(x² + x3) | b) | xn + 2 + xn + 4 = xn(x² + x4) | |
| c) | xn + 3 − xn = xn(x3 − 1) | d) | xn + 1 + xn = xn(x + 1) | |
Problem 12. In each sum, factor out the lower power of x.
For example, xn + 1 + xn + 2 = xn + 1(1 + x),
where xn + 1 is the lower power. On multiplying out, we would add the exponents (Lesson 13) and obtain the left-hand side.
| a) | xn + 4 + xn + 1 = xn + 1(x3 + 1) | b) | xn + 2 + xn + 3 = xn + 2(1 + x) | |
| c) | xn − xn − 2 = xn − 2(x² − 1) | d) | xn − 1 − xn + 1 = xn − 1(1 − x2) | |
Problem 13. The Rule of Signs. By applying the definition of the negative of a number, prove that, if ab is positive, then (−a)b is the negative of ab. That is, prove: Unlike signs produce a negative number:
(−a)b = −ab.
ab + (−a)b = [a + (−a)]b = 0· b = 0.
Therefore, since a number has one and only one negative,
(−a)b = −ab.
Example 4. x(x + 5) + 3(x + 5).
What is the common factor?
(x + 5) is the common factor. Therefore,
x(x + 5) + 3(x + 5) = (x + 3)(x + 5)
This is similar to adding like terms. In the first term, x is the coefficient of (x + 5). In the second term, 3 is its coefficient. We add the coefficients of (x + 5). And we preserve the common factor on the right.
Problem 14. Add the common factors. Do not remove parentheses.
a) x(x + 1) + 2(x + 1) = (x + 2)(x + 1)
b) x(x − 2) − 3(x − 2) = (x − 3)(x − 2)
c) x(x + 1) − (x + 1) = (x − 1)(x + 1)
d) x²(x − 5) + 4(x − 5) = (x² + 4)(x − 5)
Example 5. Factoring by grouping. Factor x3 −5x² + 3x − 15.
Solution. Group the first and second terms -- find their common factor. Do the same with the third and fourth terms.
| x3 − 5x² + 3x − 15 | = | x²(x − 5) + 3(x − 5) |
| = | (x² + 3)(x − 5). | |
Problem 15. Factor by grouping.
| a) x3 + x² + 3x + 3 | = | x²(x + 1) + 3(x + 1) |
| = | (x² + 3)(x + 1) | |
| b) 2x3 − 6x² + 5x − 15 | = | 2x²(x − 3) + 5(x − 3) |
| = | (2x² + 5)(x − 3) | |
| c) 3x3 − 15x² − 2x + 10 | = | 3x²(x − 5) − 2(x − 5) |
| = | (3x² − 2)(x − 5) | |
| d) 12x3 + 2x² − 18x − 3 | = | 2x²(6x + 1) − 3(6x + 1) |
| = | (2x² − 3)(6x + 1) | |
| e) x3 + 2x² − x − 2 | = | x²(x + 2) − (x + 2) |
| = | (x² − 1)(x + 2) | |
| f) 12x3 − 6x² − 2x + 1 | = | 6x²(2x − 1) − (2x − 1) |
| = | (6x² − 1)(2x − 1) | |
Problem 16. Show by factoring the left-hand side:
| (1 + x)² + x(1 + x)² | = | (1 + x)3. | |
| (1 + x)² + x(1 + x)² | = | (1 + x) | (1 + x)² |
| (1 + x)² is the common factor; | |||
| = | (1 + x)3 | ||
Example 6. Solve for x (Lesson 9):
| px − q | = | rx + s |
| 1. Transpose the x's to the left and everything else to the right: | ||
| px − rx | = | s + q |
| 2. Factor: | ||
| x(p − r) | = | s + q |
| 3. Solve for x: | ||
| x | = | s + q p − r |
Problem 17. Solve for x.
| ax + bx | = | c |
| x(a + b) | = | c |
| x | = | c a + b |
Problem 18. Solve for x.
| ax + b | = | cx + d |
| ax − cx | = | d − b |
| x(a − c) | = | d − b |
| x | = | d − b a − c |
Problem 19. Solve for x.
| ax − a | = | x |
| ax − x | = | a |
| x(a − 1) | = | a |
| x | = | a a − 1 |
Next Lesson: Quadratic Trinomials
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