Questions on polynomials and the factor theorem

Question 1

Show that (x − 2) is a factor of f(x) = x³ − 4x² + x + 6.

Accepted Answer

f(2) = 8 − 16 + 2 + 6 = 0, so by the factor theorem (x − 2) is a factor. 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.

Question 2

Expand (x − 2)(x² + 3x − 1) and simplify your answer.

Accepted Answer

Multiply every term in the second bracket by x, then by −2: x³ + 3x² − x − 2x² − 6x + 2. M1 for all six products. Collecting, x³ + x² − 7x + 2. A1. Two terms by three gives six terms before collecting, so a line with four terms in it has skipped some; and the constant term is (−2)(−1) = +2, the product of the two constants, which is a check on the sign at the end.

Question 3

Factorise 2x³ + 3x² − 2x completely.

Accepted Answer

Every term contains x, so take it out first: x(2x² + 3x − 2). B1 for the common factor. The quadratic factorises as (2x − 1)(x + 2), since 2x × x = 2x², (−1)(2) = −2 and the cross terms give 4x − x = 3x. M1 for factorising the quadratic. So x(2x − 1)(x + 2). A1. Completely means three factors here, and the x taken out at the start is one of them: dividing it away and factorising only the quadratic gives a pair of brackets that multiply to the wrong expression. The three roots are 0, ½ and −2.

Question 4

Divide x³ − 4x² + x + 6 by (x − 2).

Accepted 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.

Question 5

Hence factorise f(x) = x³ − 4x² + x + 6 completely, and state its roots.

Accepted 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. The roots are asked for separately, and they are the values that kill each bracket, not the numbers inside them.

Question 6

Divide 2x³ + 3x² − 5 by (x + 2), stating the quotient and remainder.

Accepted 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.

Question 7

Show that (2x − 1) is a factor of f(x) = 2x³ + 5x² − x − 1, and express f(x) as (2x − 1) times a quadratic.

Accepted 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.

Question 8

Simplify (x² + 2x − 15)/(x² − 25), and write (3x² − 5x + 4)/(x − 2) in the form ax + b + c/(x − 2).

Accepted Answer

Factorise top and bottom before cancelling anything: (x + 5)(x − 3) over (x + 5)(x − 5), and the shared bracket cancels to leave (x − 3)/(x − 5). M1 for both factorisations, A1 for the simplified fraction. Only whole factors cancel: crossing out the x² terms, or the 15 against the 25, changes the value. The second fraction is improper, so divide: 3x² − 5x + 4 = (x − 2)(3x + 1) + 6, giving 3x + 1 + 6/(x − 2). M1 for the division, A1 for all three of a = 3, b = 1 and c = 6. One expansion checks it, since (x − 2)(3x + 1) = 3x² − 5x − 2 and the 6 restores the constant.

Question 9

Factorise f(x) = x³ − 7x + 6 completely, showing how you find a first factor.

Accepted Answer

Any integer root divides the constant term 6, so the candidates are ±1, ±2, ±3 and ±6. Try the smallest: f(1) = 1 − 7 + 6 = 0, so by the factor theorem (x − 1) is a factor. M1 for testing a divisor of 6, A1 for f(1) = 0 with the theorem named. Dividing, and writing the missing x² term as 0x², gives x³ + 0x² − 7x + 6 = (x − 1)(x² + x − 6). M1 for the division with the zero term in place. The quotient factorises as (x + 3)(x − 2), so f(x) = (x − 1)(x + 3)(x − 2). A1. The other two roots, −3 and 2, also divide 6, as the rule says they must. Expanding checks it: (x − 1)(x + 3) = x² + 2x − 3, and multiplying by (x − 2) gives x³ − 7x + 6 with the x² term cancelling.

Question 10

Express 3/(x + 2) − 2/(x − 1) as a single fraction in its simplest form.

Accepted Answer

Put both over the common denominator (x + 2)(x − 1): the first fraction becomes 3(x − 1) over it and the second 2(x + 2). M1 for the common denominator with both numerators adjusted. The numerator is 3(x − 1) − 2(x + 2) = 3x − 3 − 2x − 4 = x − 7. A1. So the expression is (x − 7)/((x + 2)(x − 1)). A1. The minus sign applies to all of 2(x + 2), which is where the −4 comes from; writing −2x + 4 gives x + 1 on top instead. Nothing cancels in the true answer, since x − 7 shares no factor with either bracket.

Question 11

Given that (x − 2) is a factor of f(x) = x³ + kx² − 8x + 12, find the value of k, and factorise f(x) completely.

Accepted Answer

A factor (x − 2) means f(2) = 0: 8 + 4k − 16 + 12 = 4k + 4 = 0, so k = −1. M1 for f(2) = 0, A1 for k. Then f(x) = x³ − x² − 8x + 12, and dividing by (x − 2) gives the quotient x² + x − 6. M1 for the division. That factorises as (x + 3)(x − 2), so f(x) = (x − 2)²(x + 3). A1. The bracket (x − 2) appears twice, so the graph touches the x-axis at x = 2 rather than crossing it, and f has only two distinct roots, 2 and −3. Substituting x = −2 instead of 2, from reading the factor's sign backwards, gives −8 + 4k + 16 + 12 = 0 and k = −5, for which (x − 2) is not a factor at all: f(2) would then be 8 − 20 − 16 + 12 = −16.

Question 12

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.

Accepted 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. The last bracket can be read off the constant term rather than divided out, which saves a page of long division.

Question 13

A student factorises a cubic as (x + 3)(x² − x + 2) and keeps hunting for two more linear brackets. Explain why the hunt cannot succeed, and why every cubic nevertheless has at least one real root.

Accepted Answer

Check the quadratic before hunting further: its discriminant is (−1)² − 4(1)(2) = −7, which is negative, so x² − x + 2 has no real roots and splits into no real linear brackets. B1 for the discriminant used as the stopping rule. The factorisation (x + 3)(x² − x + 2) is therefore as far as real factors go, and the cubic has exactly one real root, at −3; more brackets would mean more roots, and there are none to find. B1 for the conclusion. A cubic can never have no real roots at all: its graph runs from one infinity to the other, so it must cross the axis at least once. B1 for the odd-degree argument.

Question 14

Given that f(x) = x³ − 3x + 2, factorise f(x) completely. Hence prove that f(x) ≥ 0 for all x ≥ −2, and state the values of x in that range for which f(x) = 0.

Accepted Answer

f(1) = 1 − 3 + 2 = 0, so (x − 1) is a factor. M1 for finding a root by the factor theorem. Dividing, with the zero x² term written in, gives f(x) = (x − 1)(x² + x − 2), and the quotient is (x + 2)(x − 1), so f(x) = (x − 1)²(x + 2). M1 for the division, A1 for the factorisation with the repeated bracket. Now argue from the factors. (x − 1)² is a square, so it is never negative for any x, and for x ≥ −2 the bracket (x + 2) is at least 0. M1 for the sign of each factor. A product of two non-negative quantities is non-negative, so f(x) ≥ 0 throughout x ≥ −2. A1 for the conclusion, with equality exactly where a factor vanishes: x = −2 and x = 1. The graph touches the axis at x = 1 and crosses at x = −2, and for x below −2 the second bracket is negative so f is negative there, which is why the range in the question stops at −2.

Question 15

Simplify (2x³ + 3x² − 8x + 3)/(2x² − 5x + 3) as far as possible, and then write the simplified expression in the form ax + b + c/(2x − 3), where a, b and c are integers.

Accepted Answer

Factorise the denominator: 2x² − 5x + 3 = (2x − 3)(x − 1). M1 for that. The numerator must share a factor for anything to cancel, so test x = 1 in it: 2 + 3 − 8 + 3 = 0, and (x − 1) is a factor of the numerator by the factor theorem. M1 for finding the shared factor. Dividing the cubic by (x − 1) gives 2x² + 5x − 3, so the numerator is (x − 1)(2x² + 5x − 3). A1. Cancelling the common bracket leaves (2x² + 5x − 3)/(2x − 3), valid for x ≠ 1, and the quadratic on top is (2x − 1)(x + 3), which shares nothing with 2x − 3, so that is as far as cancelling goes. A1 for the simplified fraction. It is still improper, so divide by (2x − 3): 2x² ÷ 2x = x, and x(2x − 3) = 2x² − 3x leaves 8x − 3; then 8x ÷ 2x = 4, and 4(2x − 3) = 8x − 12 leaves a remainder of 9. M1 for the division by a non-monic bracket. So the expression is x + 4 + 9/(2x − 3), with a = 1, b = 4 and c = 9. A1. Expanding checks it: (2x − 3)(x + 4) = 2x² + 5x − 12, and the 9 restores the −3. The hole at x = 1 does not appear in the final form, but the original expression is undefined there and the simplification does not change that.