1 a. P(x) is a polynomial.
If P (a) =0
State the factor theorem.
If P(a) = 0, then (x-a) is a factor of P(x)
Pretty much as it reads. The idea is you try integers repeatedly until the polynomial comes out at zero, then declare (x-integer) to be a factor. At a guess your next sheet will feature dividing a cubic to get a quadratic, which is then easier to solve.
b. Use the factor theorem to show that (x-2) is a factor of x3-x2-7x+2
(2)^3-(2)^2-7(2)+2 = 8 - 4 - 14 +2 = - 8, which is not 0, so -2 isn't a root or this polynomial. Bit unfortunate really, as it means the copy n paste hasn't gone well.
c. Substitute x=2, x=3, and x=1 and x= -1 into the polynomial P(x)= x3-4x2+x+6
8-8+2+6=8
27-36+3+6=0
1-4+1+6=4
-1-4-1+6=0
2 Express the quadratic x2-2x-3 in the form of (x+p)2+q where p and q are constants.
p =
-2/2, q is whatever is needed to take you back to -3 after multiplying out (x+p)^2
(x-1)^2 = x^2-2x+1, as +1 is four greater than -3, q needs to be -4 in order to make the expressions equal. Hence,
x2-2x-3 = (x-1)^2 - 4
a. Hence or otherwise find the coordinates of the vertex of the parabola y= x2-2x-3.
Otherwise:
The turning point is where the gradient is zero. Gradient is 2x-2, which equals zero at x=1, so the vertex is at (1, -4).
Hence:
the equation is that of x^2, that's been moved down four and to the right by one.
b. Find the roots of x2-2x-3=0 and hence sketch the parabola y= x2-2x-3.
By trial and error, the roots are -1 and 3. The turning point is at (x=1, y=-4) from symmetry then substitution.
3b. Express the following quadratics in the completed square form and solve them.
i. X2-4x-3=0
ii. X2+6x+4=0
1. (x-2)^2-7
Roots at [ 4(+/-)SQRT(16+12) ]/2 from the standard result, which doesn't factorise nicely
SQRT(28) = SQRT(4x7) = SQRT(2^2x7) = SQRT(2^2)SQRT(7)=2SQRT(7)
Roots at 2 + sqrt(7), and 2 - sqrt(7).
(x+3)^2-5
Roots at [ -6(+/-)SQRT(36-16) ]/2 from the standard result
Roots at SQRT(5) -3 and SQRT(5) + 3, working identical to the above
It might be a "known result" that the roots are at sqrt(-q) + or - the thing in the brackets, but it's not terribly obvious why this would always be the case. Memorising " -b + / - the square root of b squared - 4 a c all over 2 a" got me through gcse quadratic equations, so I'd suggest that instead of all this trial and error nonsense.
4a Find the gradient of above curve at point P by drawing a tangent to the curve at P
Unfortunately the there isn't an image, and rapidshare has reached its ten download limit. A tangent is a line which meets at the point in question, but doesn't cross the line either side. The gradient of the tangent is equal to the gradient of the curve at this point.
If this is gcse you probably haven't met differentiation. If it's AS you have. Either way, the rule of Ax^n goes to Anx^n-1 for each term (constants going to zero) gives you an equation for the gradient of the line. This is zero at turning points. Beats drawing graphs and measuring things hands down.
