FACTOR THEOREM
If p(x) is a polynomial of degree n ≥ 1 and a is any real number, then
(i) x – a is a factor of p(x), then p(a) = 0 and
(ii) p(a) = 0 then x – a is a factor of p(x).
REMARK

Ex. Use the factor theorem to determine whether x – 1 is a factor of
(a) x^{3} + 8x^{2} – 7x – 2
(b) 2√2 x^{3} + 5√2 x^{2} – 7√2
Sol. (a) Let p(x) = x^{3} + 8x^{2} – 7x – 2
By using factor theorem, (x – 1) is a factor of p(x) only when p(1) = 0
p(1) = (1)^{3} + 8(1)^{2} – 7(1) – 2 = 1 + 8 – 7 – 2 = 9 – 9 = 0
Hence (x – 1) is a factor of p(x).
(b) Let p(x) = 2√2x^{3} + 5√2x^{2} – 7√2
By using factor theorem, (x – 1) is a factor of p(x) only when p(1) = 0
p(1) = 2√2 (1)3 + 5√2 (1)^{2} – 7√2 = 2√2 + 5√2 – 7√2 = 7√2 – 7√2 = 0
Hence (x – 1) is a factor of p(x).
Ex. Without actual division, prove that the polynomial 2x^{3} + 13x^{2} + x – 70 is exactly divisible by x – 2.
Sol. The polynomial p(x) = 2x^{3} + 13x^{2} + x – 70 is exactly divisible by x – 2 means that x – 2 is a factor of
p(x) = 2x^{3} + 13x^{2} + x – 70.
Now p(2) =2(2)^{3} + 13(2)^{2} + 2 – 70 = 16 + 52 + 2 – 70 = 0
∴ By factor theorem, x – 2 is a factor of p(x)
i.e. p(x) = 2x^{3} + 13x^{2} + x – 70 is exactly divisible by x – 2.
FACTORISING OF POLYNOMIAL OF HIGHER DEGREE
Ex. Using factor theorem, factorize the polynomial x^{3} – 6x^{2} + 11x – 6.
Sol. Let f(x) = x^{3} – 6x^{2} + 11x – 6
The constant term in f(x) is equal to –6 and factors of – 6 are ±1, ±2, ±3, ±6. Putting x = 1 in f(x), we have
f(1) = (1)^{3} – 6 × (1)^{2} + 11 × 1 – 6 = 1 – 6 + 11 – 6 = 0 ∴ (x – 1) is a factor of f(x)
Similarly, x – 2 and x – 3 are factors of f(x). Since f(x) is a polynomial of degree 3. So, it can not have more than three linear factors.
Let f(x) = k (x – 1) (x – 2) (x – 3)
Then,x^{3} – 6x^{2} + 11x – 6 = k(x – 1) (x – 2) (x – 3)
Putting x = 0 on both sides, we get – 6 = k (0 – 1) (0 – 2) (0 – 3) ⇒ –6 = – 6 k ⇒ k = 1
Putting k = 1 in f(x) = k(x – 1) (x – 2) (x – 3), we get f(x) = (x – 1) (x – 2) (x – 3)
Hence, x^{3 }– 6x^{2} + 11x – 6 = (x – 1) (x – 2) (x – 3)
Ex. Using factor theorem, factorize the polynomial x^{4 }+ x^{3} – 7x^{2 }– x + 6.
Sol. Let f(x) = x^{4 }+ x^{3} – 7x^{2} – x + 6 the factors of constant term 6 are ±1, ±2, ±3 and ±6
Now, f(1) = 1 + 1 – 7 – 1 + 6 = 8 – 8 = 0
⇒ (x – 1) is a factor of f(x)
f(–1) = 1 – 1 – 7 + 1 + 6 = 8 – 8 = 0 ⇒ (x + 1) is a facor of f(x)
f(2) = 2^{4} + 2^{3} – 7 × 2^{2} – 2 + 6 = 16 + 8 – 28 – 2 + 6 = 0
⇒ x – 2 is a factor of f(x)
f(–2) = (–2)^{4} + (–2)^{3} – 7(–2)^{2} – (–2) + 6 = 16 – 8 – 28 + 2 + 6 = – 12 ≠ 0
⇒ x + 2 is not a factor of f(x)
f(–3) = (–3)^{4} + (–3)^{3} – 7(–3)^{2} – (–3) + 6 = 81 – 27 – 63 + 3 + 6 = 90 – 90 = 0
⇒ x + 3 is a factor of f(x).
Since f(x) is a polynomial of degree 4. So, it cannot have more than 4 linear factors
Thus, the factor of f(x) are (x – 1), (x + 1), (x – 2) and (x + 3).
Let f(x) = k(x – 1) (x + 1) (x – 2) (x + 3)
⇒ x^{4} + x^{3 }– 7x^{2} – x + 6 = k (x – 1) (x + 1) (x – 2) (x + 3) ... (i)
Putting k = 0 on both sides, we get
6 = k (–1) (1) (–2) (3) ⇒ 6 = 6 k ⇒ k = 1
Substituting k = 1 in (i), we get
x^{4} + x^{3} – 7x^{2} – x + 6 = (x – 1) (x + 1) (x – 2) (x + 3)
COMPETITION WINDOW ZERO OF A QUADRATIC POLYNOMIAL Some useful relations involving α and β : Ex. If α and β are the zeroes of the polynomial ax^{2} + bx + c. Find the value of (ii) α^{2} + β^{2} (i) (α – β)^{2} = (α + β)^{2} – 4αβ (ii) α^{2} + β^{2} = α^{2} + β2 + 2αβ – 2αβ = (α + β)^{2} – 2αβ Ex. If α and β are the zeroes of the quadratic polynomial ax^{2} + bx + c. Find the value of (i) α^{2} – β^{2} (ii) α^{3} + β^{3}. (i) α^{2} – β^{2} = (α + β) (α – β) (ii) α^{3} + β^{3} = (α + β) (α^{2 }+ β^{2} – αβ) = (α + β) [(α^{2} + β^{2} + 2αβ) – 3αβ] = (α + β) [(α + β)^{2} – 3αβ] =
TO FORM A QUADRATC POLYNOMIAL WITH THE GIVEN ZEROES Let the zeroes of a quadratic polynomial be α and β. Ex. Form the quadratic polynomial whose zeroes are 4 and 6. Ex. Form the quadratic polynomial whose zeroes are –3, 5 