Let p(x)=x^3+2x^2+kx+3
On dividing p(x) by x-3, the remainder is 21. Therefore,
P(3)=21
Substituting x=3 in p(x)
P(3)=3^3 +2*(3)^2+k*3+3
=27+18+3k+3
=48+3k
We know that, p(3)=21.
So, 48+3k=21
3k=21-48
3k=-27
k=-27/3 =-9
now, p(x) =x^3+2x^2-9x-18
-2 is a factor of p(x) on inspection. Therefore, divide p(x) by x+2 to find the
zeroes of the polynomial.
On dividing, we get the factors to be, (x^2-9)(x+2)
(x^2-3^2)(x+2)
Factorizing using the identity a^2-b^2=(a+b)(a-b) we get,
(x+3)(x-3)(x+2)
Therefore, the zeroes of the polynomials are -3,+3 and -2.
Your answer would be B, X+2 Y-1
Answer:
25%
Step-by-step explanation:
Its basic division really. The easiest way to think of it is how many times can 3 fit into 12 three fits into 12 four times in fraction that's 1/4 or 1 over 4. Just like the quarter (United States quarter) each one is 25 cents and you need 4 of then to make 1 US dollar or 100% so each group of three is 25 percent. This under standing could work with a variety of numbers 5 is what percent of 20 or 7 is what percent of 28 if it comes in 4s each is 25%. Hope this helps.
Answer:
y = 2x - 3
Step-by-step explanation:
Take any tow point from the line
(x₁ , y₁) = (1 , -1) & (x₂ , y₂) = (0 , -3)
m = 2 & (1, -1)
y - y₁ = m(x -x₁)
y - [-1] = 2(x - 1)
y + 1 = 2x - 2
y = 2x - 2 - 1
y = 2x - 3