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Polynomials and the factor theorem questions
A cubic can be factorised once you know how to test a bracket. The factor theorem says that if f(a) = 0 then (x − a) is a factor, algebraic division or comparing coefficients then splits off the quadratic, and a cubic with rational roots comes out as three brackets multiplied together.
6 original questions · 22 marks · the polynomials and the factor theorem notes · Algebra and functions
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Show that (x − 2) is a factor of f(x) = x3 − 4x2 + x + 6.
Worked answer
f(2) = 8 − 16 + 2 + 6 = 0, so by the factor theorem (x − 2) is a factor. Mind the sign. M1 for substituting x = 2, A1 for f(2) = 0 with the theorem named. The factor (x − 2) pairs with the root x = 2, and substituting −2 instead gives −20, which proves nothing. Name the theorem in the conclusion; a bare zero is only half an answer to a show-that.Divide x3 − 4x2 + x + 6 by (x − 2).
Worked answer
Divide the leading terms in turn. First x3 ÷ x = x2; multiply back and subtract, leaving −2x2 + x. Then −2x2 ÷ x = −2x, leaving −3x + 6, and finally −3 exactly. The quotient is x2 − 2x − 3 with remainder zero, as the factor theorem promised. M1 for the first division step, dM1 for completing the division, A1 for the quotient. A non-zero remainder here means an arithmetic slip somewhere above, so the zero is a free check on the division.Hence factorise f(x) = x3 − 4x2 + x + 6 completely, and state its roots.
Worked answer
The quotient factorises as x2 − 2x − 3 = (x − 3)(x + 1), so f(x) = (x − 2)(x − 3)(x + 1) and the roots are 2, 3 and −1. M1 for factorising the quotient, A1 for the three brackets, B1 for the roots. Completely means down to linear brackets, so stopping at (x − 2)(x2 − 2x − 3) loses the final mark. The roots are asked for separately, and they are the values that kill each bracket, not the numbers inside them.Divide 2x3 + 3x2 − 5 by (x + 2), stating the quotient and remainder.
Worked answer
Write the dividend with every power present, as 2x3 + 3x2 + 0x − 5, or the columns drift out of line. Dividing gives quotient 2x2 − x + 2 and remainder −9. M1 for the division with the missing term written in, A1 for the quotient, A1 for the remainder. Substitution confirms it, since f(−2) = −16 + 12 − 5 = −9, which is the remainder theorem doing the division without dividing. That check takes one line and catches the missing-term error before it costs three marks.Show that (2x − 1) is a factor of f(x) = 2x3 + 5x2 − x − 1, and express f(x) as (2x − 1) times a quadratic.
Worked answer
The root that kills (2x − 1) is x = ½, and f(½) = ¼ + 5/4 − ½ − 1 = 0, so the bracket is a factor. Dividing gives f(x) = (2x − 1)(x2 + 3x + 1). M1 for using x = ½, A1 for f(½) = 0, M1 for the division, A1 for the quadratic. For a factor (ax − b) the test value is b/a rather than b, so substituting x = 1 here proves nothing. The quadratic that comes out has leading coefficient 1, because the 2 has already been used up in the first bracket.Given that (x − 2) and (x + 3) are both factors of f(x) = x3 + ax2 + bx − 30, find the value of a and the value of b, and hence factorise f(x) completely.
Worked answer
Each factor supplies an equation. f(2) = 8 + 4a + 2b − 30 = 0 gives 2a + b = 11, and f(−3) = −27 + 9a − 3b − 30 = 0 gives 3a − b = 19. Adding the two removes b at once: 5a = 30, so a = 6 and then b = −1. With f(x) = x3 + 6x2 − x − 30, the two known brackets multiply to x2 + x − 6, and the constant −30 divided by −6 fixes the third bracket as (x + 5). So f(x) = (x − 2)(x + 3)(x + 5). M1 for f(2) = 0, A1 for 2a + b = 11, M1 for f(−3) = 0, A1 for 3a − b = 19, dM1 for solving the pair, A1 for a and b, A1 for the factorisation. Two factors mean two substitutions and two equations, and setting up both is where the opening marks sit. The last bracket can be read off the constant term rather than divided out, which saves a page of long division.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
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