Maths Help Please?

Question

Question 5) The polynomial x^3 + px^2 + qx + 6 is exactly divisiable by ( x+2) and (x-3). Find the values of p and q.

Question 9) If x^2 = a(x+2)^2 + b(x+2) + c for all values of x, find the values of a,b, and c.

Question 13) State the number of terms in the expaned form of:
a) (x+4)^7
b) (2x-5)^10
c)(1-4z)^14

Thanks in advance :)

Answer 1

ooo geez that makes no since to me haha!! im sorry no help i can do!! sorry!!

Answer 2

5)
I agree totally with Neo's steps and solution.
p = -2; q= -5.


9)
Here's an easier route to the same solution.

When x = -2, then:
4 = a*0 + b*0 + c.
Therefore c=4.

When x = -1, then:
1 = a*1 + b*1 + 4.
Therefore b = -3-a.

When x = 0, then:
0 = a*4 + b*2 + 4
Therefore 4a + 2*(-3-a) = -4,
and so
4a + 2(-3-a) = -4,
2a -6 = -4

Solution:
a=1; b=-4; c=4.

Double-check:
(x+2)^2 -4(x+2) + 4 = x^2 + 4x + 4 -4x -8 = x^2.


13) Any binomial raised to the nth power has n+1 terms.
a) 8.
b) 11.
c) 15.

Answer 3

#5.
p=0, q=-7

Answer 4

5) (x+2)(x-3)=x^2-x-6

since x^3 and 6 are in the answer, i multiply this by x-1

(x^2-x-6)(x-1)=x^3+6-5x-2x^2
p=-2, q=-5


9) 9a+3b+c=1 (x=1); 16a+4b+c=4 (x=2); 25a+5b+c=9 (x=3)

7a+b=3; 9a+b=5

2a=2;

a=1
b=-4
c=4