Step-by-step explanation:
The factor of the polynomial is (x-2)
We know that dividing a third degree polynomial by a first degree factor will yield a quadratic of the form ax²+bx+c and possibly a remainder r,except that in this case <em>r</em><em>=</em><em>0</em><em> </em><em>since</em> x-2 is a factor
We therefore equate
x³-4x²+3x+2=(x-2)(ax²+bx+c)+r
x³-4x²+3x+2=ax³+bx²+cx-2ax²-2bx-2c+r
We group terms on the RHS
x³-4x²+3x+2=ax³+x²(b-2a)+x(c-2b)-2c+r
By equating the terms on both sides,
ax³=x³,a=1
Next,
b-2a=-4
b-2(1)=-4
b-2=-4
b=-4+2
b=-2
Next,
c-2b=3
c-2(-2)=3
c+4=3
c=-1
Now we piece together our quotient quadratic using a=1,b=-2,c=-1 to obtain x²-2x-1
We now find the zeros of x²-2x-1
which are 1+√2 and 1-√2
The values of x are therefore 2,1+√2, and 1-√2
Note that I used this method in lieu of the famous polynomial division approach to teach someone also ;)
Answer:
(- 1, 2)
Step-by-step explanation:
Given a quadratic in standard form y = ax² + bx + c : a ≠ 0
Then the x- coordinate of the vertex is
= -
f(x) = x² + 2x + 3 ← is in standard form
with a = 1 and b = 2, hence
= - = - 1
Substitute x = - 1 into f(x) for corresponding value of y
f(- 1) = (- 1)² + 2(- 1) + 3 = 1 - 2 + 3 = 2
vertex = (- 1, 2 )
Answer:
the answer is 87
Step-by-step explanation:
Step-by-step explanation:
for A
for B
Answer:
Step-by-step explanation:
y=-x²-6x+7
=-(x²+6x+3²-3²)+7
=-(x+3)²+9+7
=-(x+3)²+16
axis of symmetry is x+3=0 or x=-3
vertex=(-3,16)