Plz solve it guys plz
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1). for x:
x - 1 = 0
x = 1
for k:
2(1)³ + 9(1)² + 1 + k = 0
2 + 9 + 1 + k = 0
k + 12 = 0
k = -12
2). x - 4 = 0 <=> x = 4
now,
putting this value in 2x³ - 3x² - 18x + a
we have
2(4)³ - 3(4)² - 18(4) + a = 0
2 × 64 - 3 × 16 - 18 × 4 + a = 0
128 - 48 - 72 + a = 0
128 - 120 + a = 0
a = -8
3). Let f(x) = ax³ + x² - 2x + 4a - 9
Given (x + 1) is a factor of f(x), so f(-1) = 0
f(x) = ax³ + x² - 2x + 4a - 9
f(- 1) = a( - 1 )³ + (- 1 )² - 2 (- 1) + 4a - 9
0 = - 1a + 1 + 2 + 4a - 9
0 = - 1a + 4a + 1 + 2 - 9
0 = 3a + 3 - 9
0 = 3a - 6
-3a = - 6
a = 6/3
a = 2
4). given ,
f(x)=x⁴-x³-11x²-x+a
g(x)=x+3
compairing x+3 with x-a
a=-3
now , for the f(x) to be exactly divisible by x+3
f(a)=0
or,(-3)⁴-(-3)³-11*(-3)²-(-3)+a =0
or,81-27-99+3+a=0
or,-42+a=0
or,a=42
so the required value of a is 42.
5). In the attachment ☺️
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