use the remainder theorem to find the remainder when P(x)=x^4-9x^3-5x^2-3x+4 is divided by x+3
-3[1 -9 -5 -3 4
-3 36 -93 288
1 -12 32 -96 292
x^3-12x^2+31x-96
x+3[x^4-9x^3-5x^2-3x+4
-(x^4+3x^3)
-12x^3-5x^2
-(-12x^3-36^2
31x^2-3x
(-31x^2+93x)
-96x+4
-(-96-288)
292
will the teacher know if i use lees ?
prolly,, just change it up a lil tho
Why did the polynomial go to therapy? Because it was having issues with divisibility! 🤣
To find the remainder when P(x) is divided by x+3, we can use the remainder theorem. According to the theorem, if we substitute the opposite sign of the divisor into the polynomial, the result will be the remainder.
So let's substitute -3 into P(x):
P(-3) = (-3)^4 - 9(-3)^3 - 5(-3)^2 - 3(-3) + 4
= 81 + 243 - 45 + 9 + 4
= 292
Therefore, the remainder when P(x) is divided by x+3 is 292.
To find the remainder when the polynomial P(x) is divided by x + 3, we can use the Remainder Theorem. According to this theorem, if we substitute the opposite of the divisor into the polynomial, the resulting value will be the remainder.
In this case, the divisor is x + 3, so its opposite is -3. Therefore, we need to find the value of P(-3) to get the remainder.
Let's calculate it step by step:
1. Substitute -3 into the polynomial P(x):
P(-3) = (-3)^4 - 9(-3)^3 - 5(-3)^2 - 3(-3) + 4
2. Simplify the expression:
P(-3) = 81 - 9(-27) - 5(9) + 9 + 4
P(-3) = 81 + 243 - 45 + 9 + 4
P(-3) = 292
Therefore, the remainder when P(x) is divided by x + 3 is 292.