Common factor: 2nd Level
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.
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).
Problem 12. In each sum, factor out the lower power of x.
For example, xn − 1 + xn = 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.
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.
Problem 15. Factor by grouping.
Problem 16. Show by factoring the left-hand side:
Problem 17. Solve for x.
Problem 18. Solve for x.
Problem 19. Solve for x.
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