Answer:
10. x=2 v x =-1 v x=-4
11. x=1
12 x=1 v x=(-1-√17)/2 v x=(-1+√17)/2
Explanation:
10.
f(x)=x³+3x²-6x-8
{-1,1,-2,2,-4,4,-8,8} one of these can be zeros of the function
f(2)=2³+3*2²-6*2-8=8+12-12-8=0
x=2 =>x-2=0
f(x)=(x-2)(x²+kx+4)
(x-2)(x²+kx+4)=x³+3x²-6x-8
x³+kx²+4x-2x²-2kx-8=x³+3x²-6x-8
x³+(k-2)x²+(4-2k)x-8=x³+3x²-6x-8
k-2=3 ∧ 4-2k=-6
k=5 ∧ -2k=-10
k=5 ∧ k=5
f(x)=(x-2)(x²+5x+4)
1)
4=1*4=2*2
1+4=5
x²+5x+4=x²+x+4x+4=x(x+1)+4(x+1)=(x+1)(x+4)
f(x)=(x-2)(x+1)(x+4)
f(x)=0
(x-2)(x+1)(x+4)=0
x-2=0 v x+1=0 v x+4=0
x=2 v x =-1 v x=-4
or 2)
f(x)=(x-2)(x²+5x+4)
f(x)=0
(x-2)(x²+5x+4)=0
x-2=0 v x²+5x+4=0
x=2 v x²+5x+4=0
Δ=b²-4ac
b=5 a=1 c=4
Δ=5²-4*1*4
Δ=9
√Δ=3
x=(-b-√Δ)/2a v x=(-b+√Δ)/2a
x=(-5-3)/2 v x=(-5+3)/2
x=-4 v x=-1
x=2 v x=-4 v x=-1
11
f(x)=x³ – x² + 9x− 9
f(x)=0
x³ – x² + 9x− 9=0
x²(x-1)+9(x-1)=0
(x-1)(x²+9)=0
x²+9>0 no solution in Real number
x-1=0
x=1
12
f(x)=x⁴-x³-5x²+9x-4
{-1,1,-2,2,-4,4} one of these can be zeros of the function
f(1)=1⁴-1³-5*1²+9*1-4=1-1-5+9-4=0
x=1
x-1=0
f(x)=(x-1)(x³+px²+qx+4)
(x-1)(x³+px²+qx+4)=x⁴-x³-5x²+9x-4
x⁴+px³+qx²+4x-x³-px²-qx-4=x⁴-x³-5x²+9x-4
x⁴+(p-1)x³+(q-p)x²+(4-q)x-4=x⁴-x³-5x²+9x-4
p-1=-1 ∧ q-p=-5 ∧ 4-q=9
p=0∧ q-0=-5 ∧ -q=5
p=xx0 ∧ q=-5 ∧ q=-5
p=0, q=-5
1)
f(x)=(x-1)(x³-5x+4)
x³-5x+4=x³-x-4x+4=x(x²-1)-4(x-1)=x(x-1)(x+1)-4(x-1)=(x-1)[x(x+1)-4]=(x-1)(x²+x-4)
f(x)=(x-1)(x-1)(x²+x-4) f(x)=0
(x-1)²(x²+x-4) =0
x-1=0 v x²+x-4=0
x=1 v x²+x-4=0
Δ=b²-4ac
b=1 a=1 c=-4
Δ=1²-4*1*(-4)
Δ=17
√Δ=√17
x=(-b-√Δ)/2a v x=(-b+√Δ)/2a
x=(-1-√17)/2 v x=(-1+√17)/2
x=1 v x=(-1-√17)/2 v x=(-1+√17)/2
or 2)
f(x)=(x-1)(x³-5x+4)
g(x)=x³-5x+4
{-1,1,-2,2,-4,4} one of these can be zeros of the function
g(1)=1³-5*1+4=1-5+4=0
x=1
x-1=0
g(x)=(x-1)(x²+rx-4)
(x-1)(x²+rx-4)=x³-5x+4
x³+rx²-4x-x²-rx-4=x³-5x+4
x³+(r-1)x²+(-4-r)x-4=x³-5x+4
r-1=0 ∧ -4-r=-5
r=1 ∧ -r=-1
r=1
g(x)=(x-1)(x²+x-4)
g(x)=0
(x-1)²(x²+x-4) =0
x-1=0 v x²+x-4=0
x=1 v x²+x-4=0
Δ=b²-4ac
b=1 a=1 c=-4
Δ=1²-4*1*(-4)
Δ=17
√Δ=√17
x=(-b-√Δ)/2a v x=(-b+√Δ)/2a
x=(-1-√17)/2 v x=(-1+√17)/2
x=1 v x=(-1-√17)/2 v x=(-1+√17)/2