12 multiple-choice questions, progressively harder.
Factor x4−1x^4 - 1x4−1 completely.
Solution
Correct answer: C
Read x4−1x^4 - 1x4−1 as (x2)2−12(x^2)^2 - 1^2(x2)2−12 and factor. The factor x2−1x^2 - 1x2−1 is again a difference of squares, so factor it too.
x4−1=(x2+1)(x2−1)=(x2+1)(x+1)(x−1)x^4 - 1 = (x^2 + 1)(x^2 - 1) = (x^2 + 1)(x + 1)(x - 1)x4−1=(x2+1)(x2−1)=(x2+1)(x+1)(x−1)
Stopping at (x2+1)(x2−1)(x^2 + 1)(x^2 - 1)(x2+1)(x2−1) is incomplete; the sum of squares x2+1x^2 + 1x2+1 stops there.
Factor 2x2−502x^2 - 502x2−50 completely.
Pull out the common factor 222 first, then factor the difference of squares.
2x2−50=2(x2−25)=2(x+5)(x−5)2x^2 - 50 = 2(x^2 - 25) = 2(x + 5)(x - 5)2x2−50=2(x2−25)=2(x+5)(x−5)
Stopping at 2(x2−25)2(x^2 - 25)2(x2−25) is incomplete.
Solve x2−9=0x^2 - 9 = 0x2−9=0.
Correct answer: D
Factor and use the zero-product property, which says a product is zero only when one of its factors is zero.
x2−9=(x+3)(x−3)=0 ⟹ x=3 or x=−3x^2 - 9 = (x + 3)(x - 3) = 0 \implies x = 3 \text{ or } x = -3x2−9=(x+3)(x−3)=0⟹x=3 or x=−3
Both roots solve the equation, so a single value is not enough.
Solve x2−49=0x^2 - 49 = 0x2−49=0.
Factor and apply the zero-product property.
x2−49=(x+7)(x−7)=0 ⟹ x=7 or x=−7x^2 - 49 = (x + 7)(x - 7) = 0 \implies x = 7 \text{ or } x = -7x2−49=(x+7)(x−7)=0⟹x=7 or x=−7
Both the positive and the negative root are solutions.
Use the difference of squares to compute 51×4951 \times 4951×49.
Correct answer: A
Write each factor as a distance from 505050: 51=50+151 = 50 + 151=50+1 and 49=50−149 = 50 - 149=50−1.
51×49=(50+1)(50−1)=502−12=2500−1=249951 \times 49 = (50 + 1)(50 - 1) = 50^2 - 1^2 = 2500 - 1 = 249951×49=(50+1)(50−1)=502−12=2500−1=2499
The difference of squares turns the product into a square minus 111.
Factor 3x3−12x3x^3 - 12x3x3−12x completely.
Pull out the common factor 3x3x3x first, then factor the difference of squares.
3x3−12x=3x(x2−4)=3x(x+2)(x−2)3x^3 - 12x = 3x(x^2 - 4) = 3x(x + 2)(x - 2)3x3−12x=3x(x2−4)=3x(x+2)(x−2)
Stopping at 3x(x2−4)3x(x^2 - 4)3x(x2−4) is incomplete.
Factor x4−81x^4 - 81x4−81 completely.
Read x4−81x^4 - 81x4−81 as (x2)2−92(x^2)^2 - 9^2(x2)2−92. The factor x2−9x^2 - 9x2−9 factors again.
x4−81=(x2+9)(x2−9)=(x2+9)(x+3)(x−3)x^4 - 81 = (x^2 + 9)(x^2 - 9) = (x^2 + 9)(x + 3)(x - 3)x4−81=(x2+9)(x2−9)=(x2+9)(x+3)(x−3)
The sum of squares x2+9x^2 + 9x2+9 stops there.
Factor 5x2−455x^2 - 455x2−45 completely.
Correct answer: B
Pull out the common factor 555 first, then factor the difference of squares.
5x2−45=5(x2−9)=5(x+3)(x−3)5x^2 - 45 = 5(x^2 - 9) = 5(x + 3)(x - 3)5x2−45=5(x2−9)=5(x+3)(x−3)
Stopping at 5(x2−9)5(x^2 - 9)5(x2−9) is incomplete.
Solve 4x2−9=04x^2 - 9 = 04x2−9=0.
Factor 4x2−94x^2 - 94x2−9 as a difference of squares, then use the zero-product property.
4x2−9=(2x+3)(2x−3)=0 ⟹ x=32 or x=−324x^2 - 9 = (2x + 3)(2x - 3) = 0 \implies x = \tfrac{3}{2} \text{ or } x = -\tfrac{3}{2}4x2−9=(2x+3)(2x−3)=0⟹x=23 or x=−23
Each factor set to zero gives one solution.
Factor x4−y4x^4 - y^4x4−y4 completely.
Read x4−y4x^4 - y^4x4−y4 as (x2)2−(y2)2(x^2)^2 - (y^2)^2(x2)2−(y2)2. The factor x2−y2x^2 - y^2x2−y2 factors again.
x4−y4=(x2+y2)(x2−y2)=(x2+y2)(x+y)(x−y)x^4 - y^4 = (x^2 + y^2)(x^2 - y^2) = (x^2 + y^2)(x + y)(x - y)x4−y4=(x2+y2)(x2−y2)=(x2+y2)(x+y)(x−y)
The sum of squares x2+y2x^2 + y^2x2+y2 does not factor further over the reals.
Which of these is a difference of squares?
A difference of squares needs two perfect-square terms and a minus sign. Only x4−9x^4 - 9x4−9 fits, since x4=(x2)2x^4 = (x^2)^2x4=(x2)2 and 9=329 = 3^29=32.
x4−9=(x2+3)(x2−3)x^4 - 9 = (x^2 + 3)(x^2 - 3)x4−9=(x2+3)(x2−3)
The others fail: 222 and x3x^3x3 are not squares, and x2+25x^2 + 25x2+25 is a sum.
For real numbers aaa and bbb, the equation a2−b2=0a^2 - b^2 = 0a2−b2=0 means which of the following?
a2−b2=(a+b)(a−b)=0 ⟹ a=b or a=−ba^2 - b^2 = (a + b)(a - b) = 0 \implies a = b \text{ or } a = -ba2−b2=(a+b)(a−b)=0⟹a=b or a=−b
The product is zero exactly when the two quantities are equal or exact opposites.
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