Solution:

The zero of `x + 2` is `–2`. Let `p(x) = x^3 + 3x^2 + 5x + 6` and `s(x) = 2x + 4`

Then, `p(–2) = (–2)^3 + 3(–2)^2 + 5(–2) + 6`

`= –8 + 12 – 10 + 6`
`= 0`

So, by the Factor Theorem, `x + 2` is a factor of `x^3 + 3x^2 + 5x + 6`.

Again, `s(–2) = 2(–2) + 4 = 0`

So, `x + 2` is a factor of `2x + 4`. In fact, you can check this without applying the Factor
Theorem, since `2x + 4 = 2(x + 2)`.

---

Solution:

As `x – 1` is a factor of `p(x) = 4x^3 + 3x^2 – 4x + k, p(1) = 0`

Now, `p(1) = 4(1)^3 + 3(1)^2 – 4(1) + k`

So, `4 + 3 – 4 + k = 0`

i.e., `k = –3`

We will now use the Factor Theorem to factorise some polynomials of degree `2` and `3`.
You are already familiar with the factorisation of a quadratic polynomial like
`x^2 + lx + m`. You had factorised it by splitting the middle term` lx` as `ax + bx` so that
`ab = m`. Then `x^2 + lx + m = (x + a) (x + b)`. We shall now try to factorise quadratic
polynomials of the type `ax^2 + bx + c`, where `a ≠ 0` and `a, b, c` are constants.

Factorisation of the polynomial `ax^2 + bx + c` by splitting the middle term is as follows:

Let its factors be` (px + q)` and `(rx + s)`. Then

`ax^2 + bx + c = (px + q) (rx + s) = pr x^2 + (ps + qr) x + qs`

Comparing the coefficients of `x^2`, we get `a = pr`.

Similarly, comparing the coefficients of `x`, we get `b = ps + qr`.

And, on comparing the constant terms, we get `c = qs`.

This shows us that `b` is the sum of two numbers `ps` and `qr`, whose product is

`(ps)(qr) = (pr)(qs) = ac`.

Therefore, to factorise `ax^2 + bx + c`, we have to write b as the sum of two
numbers whose product is `ac`. This will be clear from Example 13.

---

Solution:

(By splitting method) : If we can find two numbers `p` and `q` such that
`p + q = 17` and `pq = 6 × 5 = 30`, then we can get the factors.

So, let us look for the pairs of factors of `30`. Some are `1` and `30, 2 `and `15, 3` and `10, 5`
and `6`. Of these pairs, `2` and `15` will give us `p + q = 17`.

So, `6x^2 + 17x + 5 = 6x^2 + (2 + 15)x + 5`
`= 6x^2 + 2x + 15x + 5`

`= 2x(3x + 1) + 5(3x + 1)`

`= (3x + 1) (2x + 5)`

`text ( Solution 2 : )`(Using the Factor Theorem)

`6x^2 +17 x + 5 = 6 (x^2 +17/6 x + 5/6) = 6 p (x)` say. If `a` and `b` are the zeroes of `p(x)`, then

`6x^2 + 17x + 5 = 6(x – a) (x – b)`. So, `ab = 5/6` . Now , `p (1/2) = 1/4 +17/6 (1/2) +5/6 ≠ 0`. But

`p ( (-1)/3) = 0`. So, ` (x+ 1/3) ` is a factor of `p(x)`. Similarly, by trial, you can find that

`(x+ 5/2)` is a factor of ` p ( x) ` .

Therefore, `6x^2 + 17 x + 5 = 6 (x +1/3) ( x+5/2)`

`= 6 ( (3x+1)/3) ( (2x+5)/2) `

For the example above, the use of the splitting method appears more efficient. However,
let us consider another example.

---

Solution:

Let `p(y) = y^2 – 5y + 6`. Now, if` p(y) = (y – a) (y – b)`, you know that the
constant term will be `ab`. So, `ab = 6`. So, to look for the factors of `p(y)`, we look at the
factors of `6`.

The factors of `6` are `1, 2` and `3`.

Now, `p(2) = 22 – (5 × 2) + 6 = 0`

So, `y – 2` is a factor of `p(y)`.

Also, `p(3) = 3^2 – (5 × 3) + 6 = 0`

So, `y – 3` is also a factor of `y^2 – 5y + 6`.

Therefore, `y^2 – 5y + 6 = (y – 2)(y – 3)`

Note that `y^2 – 5y + 6` can also be factorised by splitting the middle term `–5y`.

Now, let us consider factorising cubic polynomials. Here, the splitting method will not
be appropriate to start with. We need to find at least one factor first, as you will see in
the following example.

---

Solution:

Let `p(x) = x^3 – 23x^2 + 142x – 120`

We shall now look for all the factors of `–120`. Some of these are `±1, ±2, ±3,
±4, ±5, ±6, ±8, ±10, ±12, ±15, ±20, ±24, ±30, ±60`.

By trial, we find that `p(1) = 0`. So `x – 1` is a factor of `p(x)`.

Now we see that `x^3 – 23x^2 + 142x – 120 = x^3 – x^2 – 22x^2 + 22x + 120x – 120`

`= x^2(x –1) – 22x(x – 1) + 120(x – 1)` (Why?)

`= (x – 1) (x^2 – 22x + 120)` [Taking `(x – 1)` common]

We could have also got this by dividing `p(x)` by `x – 1`.

Now `x^2 – 22x + 120` can be factorised either by splitting the middle term or by using
the Factor theorem. By splitting the middle term, we have:

`x^2 – 22x + 120 = x^2 – 12x – 10x + 120`

`= x(x – 12) – 10(x – 12)`

`= (x – 12) (x – 10)`

So, `x^3 – 23x^2 – 142x – 120 = (x – 1)(x – 10)(x – 12)`

---