Algebraic fractions: 2nd Level
Example 1. Write the missing numerator.
| 3 4 |
= | ___?__ 8x − 12 |
Answer. What times 4 produced 8x − 12? In other words, we must expect that 4, the denominator on the left, will be a factor on the right. We must expect that 4 has been multiplied.
| 3 4 |
= | ___?__ 8x − 12 |
= | ___?___ 4(2x − 3) |
The denominator 4 has been multiplied by 2x − 3; therefore, the numerator 3 also must be multiplied by 2x − 3.
| 3 4 |
= | 3(2x − 3) 4(2x − 3) |
= | 6x − 9 8x − 12 |
Example 2. Write the missing numerator.
| __5_ x − 3 |
= | ____?____ x2 − 4x + 3 |
Answer. What times x − 3 produced x2 − 4x + 3? Again, we must expect that the original denominator will be a factor on the right.
| __5_ x − 3 |
= | ____?____ x2 − 4x + 3 |
= | _____?_____ (x − 3)(x − 1) |
x − 3 has been multiplied by x − 1; therefore, 5 also must be multiplied by x − 1:
| __5_ x − 3 |
= | __5(x − 1)__ (x − 3)(x − 1) |
= | __5x − 5__ x2 − 4x + 3 |
Problem 1. Factor the new denominator, write the missing numerator.
| a) | 2 5 |
= | ___?___ 15x + 20 |
= | ___?___ 5(3x + 4) |
= | 2(3x + 4) 5(3x + 4) |
= |
_6x + 8_ 15x + 20 |
| b) | 2 x |
= | __?__ x2 − x |
= | __ ? __ x(x − 1) |
= |
2(x − 1) x(x − 1) |
= |
2x − 2 x2 − x |
| c) | _5_ 2x2 |
= | ____?___ 4x3 + 6x2 |
= | __ __?_ __ 2x2(2x + 3) |
| = |
5(2x + 3) 2x2(2x + 3) |
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| = |
10x + 15 4x3 + 6x2 |
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| d) | 4 x |
= | ___?___ 8x2 − 3x |
= | _ __?___ x(8x − 3) |
= |
4(8x − 3) x(8x − 3) |
= |
32x − 12 8x2 − 3x |
| e) | _2_ 3x4 |
= | ____?___ 6x6 + 3x4 |
= | ___ _?_ __ 3x4(2x2 + 1) |
= |
_2(2x2 + 1)_ 3x4(2x2 + 1) |
| = |
_4x2 + 2_ 6x6 + 3x4 |
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The following problems assumes skill with Quadratic Trinomials, Perfect Square Trinomials, and the Difference of Two Squares.
Problem 2. Factor the new denominator, and write the missing numerator.
| a) | __2__ x + 4 |
= | ____?____ x2 + 6x + 8 |
= |
__ 2(x + 2)__ (x + 4)(x + 2) |
= |
__2x + 4__ x2 + 6x + 8 |
| b) | x + 3 x − 2 |
= | ____?____ x2 + x − 6 |
= |
__ (x + 3)2__ (x − 2)(x + 3) |
= |
x2 + 6x + 9 x2 + x − 6 |
| c) | __1__ x − 2 |
= | __?__ x2 − 4 |
= |
___ x + 2___ (x − 2)(x + 2) |
= |
x + 2 x2 − 4 |
| d) | __x__ x + 2 |
= | ____?____ x2 + 5x + 6 |
= |
__x(x + 3)__ (x + 2)(x + 3) |
= |
__x2 + 3x_ x2 + 5x + 6 |
| e) | x + 3 x − 1 |
= | ____?____ x2 − 4x + 3 |
= |
(x + 3)(x − 3) (x − 1)(x − 3) |
= |
__x2 − 9__ x2 − 4x + 3 |
| f) | x − 5 x + 5 |
= | ___?__ x2 − 25 |
= |
__ (x − 5)2__ (x + 5)(x − 5) |
= |
x2 − 10x + 25 x2 − 25 |
| g) | 2x − 1 x − 5 |
= | ____?____ 2x2 − 9x − 5 |
= |
(2x − 1)(2x + 1) (x − 5)(2x + 1) |
= |
__4x2 − 1__ 2x2 − 9x − 5 |
Problem 3. Reduce to lowest terms.
| a) | x2 − x x5 − x3 |
= | _x(x − 1) x3(x2 − 1) |
= | _ x − 1 x2(x − 1)(x + 1) |
= | 1 x2(x + 1) |
| b) | x5 + x4 x3 + 2x2 + x |
= | x4(x + 1) x(x2 + 2x + 1) |
= | x3(x + 1) (x + 1)2 |
= | x3 x + 1 |
Problem 4. Reduce to lowest terms—if possible.
See Lesson 19, Problems 11 and 12.
| a) | x3 − 1 x − 1 |
= | (x − 1)(x2 + x + 1) x − 1 |
= | x2 + x + 1 |
| b) | a2 − b2 a3 − b3 |
= | (a − b)(a + b) (a − b)(a2 + a + 1) |
= | a + b a2 + a + 1 |
| c) | a2 + b2 a3 +b3 |
= | Not possible. a2 + b2 cannot be factored. |
The following problem depends on knowing that b − a is the negative of a − b. (Lesson 7.)
Problem 5. Simplify.
| a) | 2 − x x − 2 |
= | −1 | b) | x(x − 3) 3 − x |
= | x(−1) = −x |
| c) | __1 − x_ x2(x − 1) |
= |
1 x2 |
(−1) | = | − |
1 x2 |
| d) | 3x − 6 2 − x |
= |
3(x − 2) 2 − x |
= | 3(−1) | = | −3 |
| e) | x2 − 7x + 12 4 −x |
= |
(x − 4)(x − 3) 4 −x |
= | −1(x − 3) | = | 3 − x |
| f) | __1 − x__ x2 − 2x + 1 |
= |
___1 − x__ (x − 1)(x − 1) |
= |
−1 x − 1 |
= |
1 −(x − 1) |
= |
1 1 − x |
Next Lesson: Multiplying and dividing algebraic fractions
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