Helppp i am stuck.
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sin^2x - 0.3sinx - 0.04
= 1/100 [ 100 sin^2(x) - 30 sin(x) - 4 ]
= 1/100 [ 100sin^2(x) - 40 sin(x) + 10sin(x) - 4 ]
= 1/100 [ 20 sin x (5 sin x - 2) + 2( 5 sin x - 2) ]
= 1/100 (20 sin x + 2)(5 sin x - 2)
= (2 sin x + 0.2)(0.5 sin x - 0.2)
= 1/100 [ 100 sin^2(x) - 30 sin(x) - 4 ]
= 1/100 [ 100sin^2(x) - 40 sin(x) + 10sin(x) - 4 ]
= 1/100 [ 20 sin x (5 sin x - 2) + 2( 5 sin x - 2) ]
= 1/100 (20 sin x + 2)(5 sin x - 2)
= (2 sin x + 0.2)(0.5 sin x - 0.2)
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sin²x - 0.3sinx - 0.04 ← First, factor out 0.01 or 1/100
0.01(100sin²x - 30sinx - 4) ← We can, also, factor out 2 now
0.02(50sin²x - 15sinx - 2) ← Now, we need two factors of 50(-2), i.e. -100,
that add to -15. Those would be -20 & +5
0.02[50sin²x + 5sinx - 20sinx - 2] ← Rewrote the middle term -15sinx
using -20 & +5 as coefficients.
Now, begin factoring inside the [ ]'s.
0.02[5sinx(10sinx + 1) - 2(10sinx + 1)] ← Now, factor out (10sinx + 1)
0.02[(5sinx - 2)(10sinx + 1)]
0.02(5sinx - 2)(10sinx + 1) ← ANSWER
OR
¹∕₅₀(5sinx - 2)(10sinx + 1)
Have a good one
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sin²x - 0.3sinx - 0.04 ← First, factor out 0.01 or 1/100
0.01(100sin²x - 30sinx - 4) ← We can, also, factor out 2 now
0.02(50sin²x - 15sinx - 2) ← Now, we need two factors of 50(-2), i.e. -100,
that add to -15. Those would be -20 & +5
0.02[50sin²x + 5sinx - 20sinx - 2] ← Rewrote the middle term -15sinx
using -20 & +5 as coefficients.
Now, begin factoring inside the [ ]'s.
0.02[5sinx(10sinx + 1) - 2(10sinx + 1)] ← Now, factor out (10sinx + 1)
0.02[(5sinx - 2)(10sinx + 1)]
0.02(5sinx - 2)(10sinx + 1) ← ANSWER
OR
¹∕₅₀(5sinx - 2)(10sinx + 1)
Have a good one
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