If x² + x - 12 divides p(x) = x³ + ax² + bx - 8 exactly, find a and b
Pls answer with proper steps ...
the answer is a = 5/3 and b = -34/3
Pls ... I repeat I want the steps not the answer
Spammers will confirmly be reported
Answers
Answer:
a = 5/3 and b = -34/3
Step-by-step explanation:
p(x) = x³ + ax² + bx - 8
let g(x) = x² + x - 12
according to factor to theorem, for g(x) to be a factor of p(x), p(a) = 0, such that a is the zero of g(x)
since g(x) is a quadratic equation, it can be split into mid terms to find the zeroes
=> x² + x - 12 = 0
=> x² + 4x - 3x - 12 = 0
=> (x - 3)(x + 4) = 0
=> zeroes are 3 and -4
=> p(3) = p(-4) = 0
p(3) = 0
=> 3³ + a(3)² + b(3) - 8 = 0
=> 27 - 8 + 9a + 3b = 0
=> 9a + 3b + 19 = 0 ... (i)
p(-4) = 0
=> (-4)³ + a(-4)² + b(-4) - 8 = 0
=> -64 + 16a - 4b - 8 = 0
=> 16a - 4b - 72 = 0
=> 4a - b - 18 = 0
=> 12a - 3b - 54 = 0 ... (ii)
adding (i) and (ii),
12a + 9a + 19 - 54 = 0
=> 21a = 35
=> a = 5/3
substituting the value of a,
4(5/3) - b - 18 = 0
=> b = -34/3
answer :-
a = 5/3 and b = -34/3
Step-by-step explanation:
firstly we'll factorise x² + x - 12
= x²+4x-3x-12
= x(x+4) -3(x+4)
=(x-3)(x+4)
so x-3=0. or. x+4=0
x=3. x=-4
now , by factor theorem
p(x) =0
p(3)=3³ +a(3)²+3b-8=0
27+9a+3b=8
b = (-19-9a)/3 ------------------------eq(1)
now
f(-4) =(-4)³+a(-4)²-4b-8=0
-64+16a-4b-8=0
16a-4b = 72
4a-b=18---------------------eq(2)
now substitute b = (-19-9a)/3 in eq(2)
4a-(-19-9a)/3 =18
4a+(19+9a)/3=18
12a+9a+19 =18×3
21a =54-19
21a=35
a=35/21=5/3
now put value of a in eq (2)
4(5/3)-b=18
20/3 -b =18
-b=18-20/3
b =-34/3