O-LevelAdditional MathematicsFactors of polynomials4037/11 · May/June 2019 · Q36 Marks
The polynomial p(x) = (2x-1)(x+k)-12, where k is a constant.
- (i)Write down the value of p(-k).[1]
- (ii)When p(x) is divided by x + 3 the remainder is 23. Find the value of k.[2]
- (iii)Using your value of k, show that the equation p(x) =-25 has no real solutions.[3]
Mark scheme
Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.
- [B1] -12
- [M1] (2(-3)-1)(-3+k)-12 = 23 oe or 2(-3)²+(2k-1)(-3)-k-12 = 23
- [A1] k = -2
- [M1] expansion and simplification to a 3 term quadratic equation equated to zero, using their k.
- [M1] using discriminant for their three term quadratic equation
- [A1] cao for correct discriminant and correct conclusion
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