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Let \( f(x) = -6x^3 - 7x^2 + 40x + 21 \) - Edexcel - A-Level Maths Pure - Question 8 - 2016 - Paper 2
Question 8
Let \( f(x) = -6x^3 - 7x^2 + 40x + 21 \). (a) Use the factor theorem to show that \( (x + 3) \) is a factor of \( f(x) \). (b) Factorise \( f(x) \) completely. (c) ... show full transcript
Worked Solution & Example Answer:Let \( f(x) = -6x^3 - 7x^2 + 40x + 21 \) - Edexcel - A-Level Maths Pure - Question 8 - 2016 - Paper 2
Step 1
Use the factor theorem to show that (x + 3) is a factor of f(x)
Answer
To show that ( (x + 3) ) is a factor of ( f(x) ), we substitute ( x = -3 ) into the function:
[ f(-3) = -6(-3)^3 - 7(-3)^2 + 40(-3) + 21 ] Calculating each term:
- ( -6(-27) = 162 )
- ( -7(9) = -63 )
- ( -120 )
- ( +21 )
Combining these: [ f(-3) = 162 - 63 - 120 + 21 = 0 ] Since ( f(-3) = 0 ), by the factor theorem, ( (x + 3) ) is indeed a factor.
Step 2
Factorise f(x) completely
Answer
Using synthetic division or polynomial long division, we divide ( f(x) ) by ( (x + 3) ):
After division, we find: [ f(x) = (x + 3)(-6x^2 + 11x + 7) ]
Next, we factor ( -6x^2 + 11x + 7 ) completely. To do this, we look for two numbers that multiply to ( -42 ) (the product of -6 and 7) and add to ( 11 ). The numbers 14 and -3 work:
So we can express this as: [ -6x^2 + 14x - 3x + 7 = -2x(3x - 7) + 1(3x - 7) = (3x - 7)(-2x + 1) ]
Therefore: [ f(x) = (x + 3)(3x - 7)(-2x + 1) ]
Step 3
Hence solve the equation 6(2^y) + 7(2^{2y}) = 40(2^y) + 21
Answer
First, we rewrite the equation: [ 6(2^y) + 7(2^{2y}) - 40(2^y) - 21 = 0 ]
Letting ( x = 2^y ), we can rewrite it as: [ 7x^2 - 34x - 21 = 0 ]
Now, using the quadratic formula, ( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ): Where ( a = 7, b = -34, c = -21 ) We find: [ b^2 - 4ac = 34^2 - 4(7)(-21) = 1156 + 588 = 1744 ] Now, calculate: [ x = \frac{34 \pm \sqrt{1744}}{14} ] [ x = \frac{34 \pm 41.7}{14} ] Which gives us two possible values for ( x ): [ x_1 \approx 5.5, ; x_2 \approx -0.5 \text{ (not valid as } x = 2^y) ]
Thus, we have ( x = 2^y = 5.5 ) or ( 2^y = 5.5 ), giving: [ y = \log_2(5.5) \approx 2.32 ]
Therefore, rounding to two decimal places: [ y \approx 2.32 ]