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secondary 3 | A Maths
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Date Posted: 1 year ago
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i)
f(x) = 2x³+ax²+x+b
2x²+x–1 = (2x–1)(x+1)
Therefore 2x–1 and x+1 are factors of f(x)
By factor theorem,
When x = ½, f(x) = 0
0 = 2(½)³+a(½)²+½+b
0 = ¾+¼a+b → (1)
When x = –1, f(x) = 0
0 = 2(–1)³+a(–1)²–1+b
0 = –3+a+b
b = 3–a → (2)
Sub (2) into (1)
0 = ¾+¼a+3–a
a = 5
Sub a = 5 into (2)
b = 3–5 = –2
ii)
f(x) = 2x³+5x²+x–2 = (2x–1)(x+1)(x+2)
When f(x) = 0
0 = (2x–1)(x+1)(x+2)
2x–1 = 0 or x+1 = 0 or x+2 = 0
x = ½ or x = –1 or x = –2
iii)
¼x³+5/4x²+½x–2 = 0
2(½x)³+5(½x)²+(½x)–2 = 0
Therefore
½x = ½ or ½x = –1 or ½x = –2
x = 1 or x = –2 or x = –4
f(x) = 2x³+ax²+x+b
2x²+x–1 = (2x–1)(x+1)
Therefore 2x–1 and x+1 are factors of f(x)
By factor theorem,
When x = ½, f(x) = 0
0 = 2(½)³+a(½)²+½+b
0 = ¾+¼a+b → (1)
When x = –1, f(x) = 0
0 = 2(–1)³+a(–1)²–1+b
0 = –3+a+b
b = 3–a → (2)
Sub (2) into (1)
0 = ¾+¼a+3–a
a = 5
Sub a = 5 into (2)
b = 3–5 = –2
ii)
f(x) = 2x³+5x²+x–2 = (2x–1)(x+1)(x+2)
When f(x) = 0
0 = (2x–1)(x+1)(x+2)
2x–1 = 0 or x+1 = 0 or x+2 = 0
x = ½ or x = –1 or x = –2
iii)
¼x³+5/4x²+½x–2 = 0
2(½x)³+5(½x)²+(½x)–2 = 0
Therefore
½x = ½ or ½x = –1 or ½x = –2
x = 1 or x = –2 or x = –4
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