find the remainder in each of the following cases (i)when (2x^4-6x^3+2x^2-x+2)is divided by x+1
Answers
Answer:
What will be the remainder when f(x) = 2x^4 - 6x^3 + 2x^2-x+2 is divided by g(x) =x+2?
Given, f(x) = 2x^4 -6x^3 +2x^2 -x+2 & g(x) = x+2.
Now, suppose, f(x) = g(x)×q(x)+r(x) [Remainder Theorem.]; where, q(x) is the respective quotient when f(x) is divided by g(x) & r(x) is the corresponding remainder.
Putting x = -2, we get,
f(-2) = g(-2)×q(-2)+r(-2)
=> 2(-2)^4 -6(-2)^3 +2(-2)^2 -(-2)+2 = 0+r(-2) [since, g(-2) = 0.]
=> r(-2) = 32+48+8+4
i.e., r(-2) = 92.
Hence, the required remainder is 92. (Ans)
Hope, it'll help..!!
P.S.- Feel free to comment if you have any query with it.
Thank You!
What is the remainder when 2x^4-6x^3+2x^2-x+2/x+2?
Find the remainder when the polynomial f(x) 2x^4-6x^3+2x^2-x+2 is divided by x+2?
What is the remainder when (2x^4-12x^3-13x^2+4x-9) is divided by (x+7)?
What is the quotient and remainder when f(x) =4x^3+3x^2-2x+1 is divided by 2x+1?
When x^3+2x^2-x-2 is divided by x+4, what is the remainder?
We want the remainder when f(x)=2x4−6x3+2x2−x+2 is divided by x+2.
The remainder when f(x) is divided by (x−a) is f(a).
In t his case x−a=x+2⇒a=−2.
f(−2)=2(−2)4−6(−2)3+2(−2)2−(−2)+2=92.
⇒ The remainder when f(x)=2x4−6x3+2x2−x+2 is divided by x+2 is 92.
f(x)=2x4−6x3+2x2−x+2
=2x3(x+2)−10x2(x+2)+22x(x+2)−45(x+2)+92
=(x+2)(2x3−10x2+22x−45)+92
=g(x)∗Q(x)+R where the quotient Q(x)=2x3−10x2+22x and the remainder R=92
What is remainder if 2x^2-6x^3+2x^2-x+2/x+2?
What is the remainder when p(x) =x^4-3x^3-2x^2-1, is divided by x-2?
What is the remainder when 5x^3+2x^2-7x-5 is divided by x-2?
What is the remainder when x^2-3x+2 is divided by x-2?
What is the remainder when 2x^2+3x+1 is divided by (x-1?
What is the remainder when 2x^4-6x^3+2x^2-x+2/x+2?
Soln: We have various methods to solve this problem like common division, remainder theorem, factor theorem, synthetic division. But, very easy method in these are synthetic division, remainder theorem and factor theorem that according to me. So, now I'll explain in the reminder theorem.
So, let x+2=0.
=> x = 0–2
x = -2
Therefore, we get the value of 'x', that is -2. So, now substitute the -2 in the given equation.
f(x) = 2x⁴-6x³+2x²-x+2
=> f(-2) = 2(-2)⁴-6(-2)³+2(-2)²-(-2)+2
f(-2) = 2(16)-6(-8)+2(4)+2+2
f(-2) = 32+48+8+2+2
f(-2) = 92
Therefore, 92 will be the remainder for the given equation. So, 92 is the answer for your problem. Thank You.
If f(x) =x^2 +2x+3 find 'x' if f(x) =f (2x-1)?
f(x)=x^2+2x+3
f(2x-1)= (2x-1)^2 +2(2x-1)+3
f(x) = f(2x-1)
x^2+2x+3 =(2x-1)^2+2(2x-1)+3
or x^2+2x =4x^2–4x+1+4x-2
or x^2+2x=4x^2–1
or 3x^2–2x-1 =0
or 3x^2–3x+x-1=0
or 3x(x-1)+1(x-1)=0
or (x-1) (3x+1) =0
x = 1 , -1/3 , Answer.
How do I solve? 2^2x+3 (2^x)-4=0
Let y=2x and rewrite the problem as:
y2+3y−4=0
Factorise:
(y+4)(y−1)=0
This gives us y=−4,y=1.
Take our y=2x from earlier and solve the substitution:
i) y=−4,y=2x
2x=−4 Has no solution.
ii) y=1,y=2x
2x=1 Has solution x=0 .
So our assumed solution is x=0 .
Check that it works with our starting equation:
22x+3(2x)−4=0
20+3(20)−4=0
1+3−4=0