i understand the concepts and all the theorems for this topic, i am able to do some questions, but have difficulty with others, and also factorising cubic equations from scratch. thanks for helping will +reps for everyone, over a period of a few days(:
1. Given that f(x)=x^2n - (p+1)x^2 +p where n and p are positive. Show that (x+1) is a factor of f(x) for all values of p. When p=4, find the value of n for which (x+2) is a factor of f(x) and, for this case, factorise f(x) completely.
2. The expression 2x^3 - 3x^2 +ax +2 leaves a remainder of R when divided by 2x+1 and a remainder of 4R+5 when divided by x-3. Find the value of a.
3. Factorise the expression 6x^3 - 7x^2 - x +2 fully. Hence find all the factors of 6(y+1)^3 - 7(y+1)^2 -y+1,
solve the equation 3(2)^3p+1 + 2 = 7(2)^2p + 2^2p, giving the values of p to two decimal places where necessary.
4. find the value of p for which (2+p)x + 6px^2 + p^2 -4p is divisible by x-1 but not x+1.
You titled this "remainder and factor theorem" so presumably you know that the remainder when polynomial P(x) is divided by x-a is P(a). In particular, x-a is a factor of P(x) if and only if P(a)= 0.
that is enough to tell you that x-(-1)= x+ 1 is a factor. If p= 4, this is . x+2 is a factor if and only if x= -2 satisfies the equation . That is, or . That should be easy to solve.
2x+ 1= 2(x+ 1/2)= 2(x-(-1/2)) leaves remainder R so P(x) divided by x+1/2 would leave remainder 2R. P(-1/2)= -2/8- 3/4- (1/2) a+ 2= (1/2)a- (3/2)= R or, multiplying by 4, 2a- 6= 4R. Similarly, since P(x) divided by x-3 leaves remainder 4R+ 2, we have P(3)= 2(3^3)- 3(3^2)+ 3a+ 2= 3a+ 29= 4R+5 or 3a- 5= 4R. 4R= 2a- 6= a- 5 and now you can solve for a.2. The expression 2x^3 - 3x^2 +ax +2 leaves a remainder of R when divided by 2x+1 and a remainder of 4R+5 when divided by x-3. Find the value of a.
Once more, it is easy to see that [tex]6(1^3)- 7(1^2)- (1)+ 2= 6- 7- 1+ 2= 0 so x-1 is a factor. Factoring out x- 1, it is easy to see that the leading term of the other factor must be and the constant term3. Factorise the expression 6x^3 - 7x^2 - x +2 fully. Hence find all the factors of 6(y+1)^3 - 7(y+1)^2 -y+1,
solve the equation 3(2)^3p+1 + 2 = 7(2)^2p + 2^2p, giving the values of p to two decimal places where necessary.
must be -2. That leaves only the coefficient of x to be determined: [tex](x-1)(6x^2+ ?x- 2)= 6x^3- (6- ?)x^2- (?+2)x+ 2. So we must have 6- ?= 7 and ?+ 2= 1 so both equations give ?= -1.
. You can use the quadratic formula to find the factors of .
If it is divisible by x- 1 than so . p is either -2 or -1. If p were equal to -1, x= -1 would also make that polynomial 0 so x+ 1 would be a factor. Since it is not, p must be -2.4. find the value of p for which (2+p)x + 6px^2 + p^2 -4p is divisible by x-1 but not x+1.