Photo AI
f(x) = -6x^3 - 7x^2 + 40x + 21 (a) Use the factor theorem to show that (x + 3) is a factor of f(x) - Edexcel - A-Level Maths Pure - Question 7 - 2016 - Paper 2
Question 7
f(x) = -6x^3 - 7x^2 + 40x + 21 (a) Use the factor theorem to show that (x + 3) is a factor of f(x). (b) Factorise f(x) completely. (c) Hence solve the equation 6... show full transcript
Worked Solution & Example Answer:f(x) = -6x^3 - 7x^2 + 40x + 21 (a) Use the factor theorem to show that (x + 3) is a factor of f(x) - Edexcel - A-Level Maths Pure - Question 7 - 2016 - Paper 2
Step 1
Use the factor theorem to show that (x + 3) is a factor of f(x)
Answer
To apply the factor theorem, we substitute ( x = -3 ) into the function:
[ f(-3) = -6(-3)^3 - 7(-3)^2 + 40(-3) + 21 ]
Calculating this, we get:
[ f(-3) = -6(-27) - 7(9) - 120 + 21 ]
[ = 162 - 63 - 120 + 21 = 0 ]
Since ( f(-3) = 0 ), it follows that ( (x + 3) ) is indeed a factor of ( f(x) ).
Step 2
Factorise f(x) completely
Answer
We start with ( f(x) = -6x^3 - 7x^2 + 40x + 21 ) and already established that ( (x + 3) ) is a factor.
Using polynomial division or synthetic division, we divide ( f(x) ) by ( (x + 3) ):
- Result gives us ( -6x^2 + 11 ) as a quotient.
- Then we factor ( -6x^2 + 11 ) further, leading to ( -1(6x^2 - 11) = -1(3x -\sqrt{11})(3x + \sqrt{11}) ).
So, the complete factorization is:
[ f(x) = -1 (x + 3)(3x - \sqrt{11})(3x + \sqrt{11}) ]
Step 3
Hence solve the equation 6(2^y) + 7(2^{2y}) = 40(2^y) + 21
Answer
Let ( z = 2^y ). Then the equation becomes:
[ 6z + 7z^2 = 40z + 21 ]
Rearranging gives:
[ 7z^2 - 34z - 21 = 0 ]
Using the quadratic formula ( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ):
Where ( a = 7, b = -34, c = -21 ) yields:
[ z = \frac{34 \pm \sqrt{(-34)^2 - 4(7)(-21)}}{2(7)} ]
Calculating gives two possible z-values. Pick the valid one and then find ( y ) using ( y = \log_2(z) ).
Finally, rounding to two decimal places gives ( y \approx 1.22 ).