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, the leading term in 3x². We like the leading term to be positive. 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 + 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.
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 13. See the common factor. 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 14. Factor by grouping.
Problem 15. Show by factoring the left-hand side:
Example 6. Solve for x (Lesson 9):
Problem 16. Solve for x.
Problem 17. Solve for x.
Problem 18. Solve for x.
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