Advanced Grade 9 Polynomials worksheet covering the Factor Theorem with 12 problems on polynomial factorization and value determination.

Let's solve each problem step-by-step using the Factor Theorem and other algebraic techniques. The Factor Theorem states:

> If $ (x - a) $ is a factor of a polynomial $ p(x) $, then $ p(a) = 0 $.
> Conversely, if $ p(a) = 0 $, then $ (x - a) $ is a factor of $ p(x) $.

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1. Find the value of 'a' for which (x−1) is a factor of the polynomial $ a^2x^3 - 4ax + 4a - 1 $.

Let $ p(x) = a^2x^3 - 4ax + 4a - 1 $

Since $ (x - 1) $ is a factor, $ p(1) = 0 $

$$
p(1) = a^2(1)^3 - 4a(1) + 4a - 1 = a^2 - 4a + 4a - 1 = a^2 - 1
$$

Set $ p(1) = 0 $:
$$
a^2 - 1 = 0 \Rightarrow a^2 = 1 \Rightarrow a = \pm 1
$$

✔ Answer: $ a = 1 $ or $ a = -1 $

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2. For what value of k is the polynomial $ p(x) = 2x^3 - kx^2 + 3x + 10 $ exactly divisible by $ (x+2) $?

Since $ (x+2) $ is a factor, $ p(-2) = 0 $

$$
p(-2) = 2(-2)^3 - k(-2)^2 + 3(-2) + 10 = 2(-8) - k(4) -6 + 10 = -16 - 4k + 4 = -12 - 4k
$$

Set $ p(-2) = 0 $:
$$
-12 - 4k = 0 \Rightarrow -4k = 12 \Rightarrow k = -3
$$

✔ Answer: $ k = -3 $

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3. Find the value of 'k' if $ (x-1) $ is a factor of $ p(x) = 2x^2 + kx + \sqrt{2} $

Use $ p(1) = 0 $

$$
p(1) = 2(1)^2 + k(1) + \sqrt{2} = 2 + k + \sqrt{2}
$$

Set equal to zero:
$$
2 + k + \sqrt{2} = 0 \Rightarrow k = -2 - \sqrt{2}
$$

✔ Answer: $ k = -2 - \sqrt{2} $

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4. Factorize: $ 6x^2 + 7x - 3 $

We look for two numbers whose product is $ 6 \times (-3) = -18 $ and sum is 7.

Such numbers: $ 9 $ and $ -2 $

Split middle term:
$$
6x^2 + 9x - 2x - 3 = 3x(2x + 3) -1(2x + 3) = (3x - 1)(2x + 3)
$$

✔ Answer: $ (3x - 1)(2x + 3) $

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5. Factorize: $ (x + y)^3 - (x^3 + y^3) $

First expand $ (x + y)^3 $:
$$
(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3
$$

Now subtract $ x^3 + y^3 $:
$$
(x^3 + 3x^2y + 3xy^2 + y^3) - (x^3 + y^3) = 3x^2y + 3xy^2 = 3xy(x + y)
$$

✔ Answer: $ 3xy(x + y) $

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6. Using factor theorem, show that $ (m-n), (n-p), (p-m) $ are factors of $ m(n^2 - p^2) + n(p^2 - m^2) + p(m^2 - n^2) $

Let $ f(m, n, p) = m(n^2 - p^2) + n(p^2 - m^2) + p(m^2 - n^2) $

We want to show that $ (m - n) $, $ (n - p) $, $ (p - m) $ are factors.

Let’s simplify $ f(m,n,p) $:

$$
f = m(n^2 - p^2) + n(p^2 - m^2) + p(m^2 - n^2)
= mn^2 - mp^2 + np^2 - nm^2 + pm^2 - pn^2
$$

Group terms:
$$
= (mn^2 - pn^2) + (-mp^2 + np^2) + (-nm^2 + pm^2)
= n^2(m - p) + p^2(n - m) + m^2(p - n)
$$

Alternatively, let's try plugging in values.

#### Show $ (m - n) $ is a factor:

Let $ m = n $. Then substitute into $ f $:

$$
f(n, n, p) = n(n^2 - p^2) + n(p^2 - n^2) + p(n^2 - n^2) = n(n^2 - p^2) + n(p^2 - n^2) + 0
= n(n^2 - p^2) - n(n^2 - p^2) = 0
$$

So $ f = 0 $ when $ m = n $ → $ (m - n) $ is a factor.

Similarly, set $ n = p $:
$$
f(m, p, p) = m(p^2 - p^2) + p(p^2 - m^2) + p(m^2 - p^2) = 0 + p(p^2 - m^2) + p(m^2 - p^2) = p(p^2 - m^2) - p(p^2 - m^2) = 0
$$

So $ (n - p) $ is a factor.

Set $ p = m $:
$$
f(m, n, m) = m(n^2 - m^2) + n(m^2 - m^2) + m(m^2 - n^2) = m(n^2 - m^2) + 0 + m(m^2 - n^2)
= m(n^2 - m^2) - m(n^2 - m^2) = 0
$$

So $ (p - m) $ is a factor.

✔ Conclusion: $ (m - n), (n - p), (p - m) $ are all factors.

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7. Factorize: $ 9x^3 - 3x^2 - 5x - 1 $

Try Rational Root Theorem: possible rational roots are $ \pm1, \pm\frac{1}{3}, \pm\frac{1}{9} $

Test $ x = 1 $:
$$
9(1)^3 - 3(1)^2 - 5(1) - 1 = 9 - 3 - 5 - 1 = 0 \Rightarrow x = 1 $ is a root.
$$

So $ (x - 1) $ is a factor.

Use synthetic division:

```
1 | 9 -3 -5 -1
| 9 6 1
-------------------
9 6 1 0
```

Quotient: $ 9x^2 + 6x + 1 $

Factor $ 9x^2 + 6x + 1 = (3x + 1)^2 $

So full factorization:
$$
(x - 1)(3x + 1)^2
$$

✔ Answer: $ (x - 1)(3x + 1)^2 $

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8. Factorize: $ x^3 + 2x^2 - 5x - 6 $

Try rational roots: $ \pm1, \pm2, \pm3, \pm6 $

Test $ x = 1 $: $ 1 + 2 - 5 - 6 = -8 $ ✘
$ x = -1 $: $ -1 + 2 + 5 - 6 = 0 $ ✔

So $ (x + 1) $ is a factor.

Synthetic division:

```
-1 | 1 2 -5 -6
| -1 -1 6
------------------
1 1 -6 0
```

Quotient: $ x^2 + x - 6 = (x + 3)(x - 2) $

So full factorization:
$$
(x + 1)(x + 3)(x - 2)
$$

✔ Answer: $ (x + 1)(x + 3)(x - 2) $

---

9. Factorize: $ x^3 - 12x^2 + 47x - 60 $

Try rational roots: $ \pm1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 $

Try $ x = 1 $: $ 1 - 12 + 47 - 60 = -24 $ ✘
$ x = 2 $: $ 8 - 48 + 94 - 60 = 8 - 48 = -40; +94 = 54; -60 = -6 $ ✘
$ x = 3 $: $ 27 - 108 + 141 - 60 = (27 - 108) = -81; +141 = 60; -60 = 0 ✔

So $ x = 3 $ is a root → $ (x - 3) $ is a factor.

Synthetic division:

```
3 | 1 -12 47 -60
| 3 -9 60
---------------------
1 -9 38 0
```

Quotient: $ x^2 - 9x + 38 $

Discriminant: $ 81 - 152 = -71 < 0 $ → no real roots.

But wait — maybe I made a mistake?

Wait: $ x = 3 $: $ 27 - 108 = -81; +141 = 60; -60 = 0 $ → correct.

But $ x^2 - 9x + 38 $ has no real roots. But perhaps we can factor further?

Wait — try $ x = 4 $: $ 64 - 192 + 188 - 60 = (64 - 192) = -128; +188 = 60; -60 = 0 → yes!

Wait — $ x = 4 $: $ 4^3 = 64 $, $ 12×16 = 192 $? No: $ 12×(4)^2 = 12×16 = 192 $? Wait:

Wait: $ x = 4 $: $ 64 - 12×16 + 47×4 - 60 = 64 - 192 + 188 - 60 $

Calculate:
- $ 64 - 192 = -128 $
- $ -128 + 188 = 60 $
- $ 60 - 60 = 0 $ ✔

So $ x = 4 $ is also a root.

But earlier we had $ x = 3 $ as root.

Try $ x = 5 $: $ 125 - 300 + 235 - 60 = (125 - 300) = -175; +235 = 60; -60 = 0 ✔

So $ x = 5 $ is also a root.

So roots are $ x = 3, 4, 5 $

So factorization: $ (x - 3)(x - 4)(x - 5) $

Check: $ (x - 3)(x - 4) = x^2 - 7x + 12 $, times $ (x - 5) $:

$$
(x^2 - 7x + 12)(x - 5) = x^3 - 5x^2 - 7x^2 + 35x + 12x - 60 = x^3 - 12x^2 + 47x - 60
$$

Perfect.

✔ Answer: $ (x - 3)(x - 4)(x - 5) $

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10. State Factor Theorem. Using Factor Theorem, factorize $ x^3 - 3x^2 - x + 3 $

Factor Theorem:
If $ (x - a) $ is a factor of a polynomial $ p(x) $, then $ p(a) = 0 $.
Conversely, if $ p(a) = 0 $, then $ (x - a) $ is a factor.

Now factor $ p(x) = x^3 - 3x^2 - x + 3 $

Try $ x = 1 $: $ 1 - 3 - 1 + 3 = 0 $ ✔ → $ (x - 1) $ is a factor

Synthetic division:

```
1 | 1 -3 -1 3
| 1 -2 -3
-------------------
1 -2 -3 0
```

Quotient: $ x^2 - 2x - 3 = (x - 3)(x + 1) $

So full factorization:
$$
(x - 1)(x - 3)(x + 1)
$$

✔ Answer: $ (x - 1)(x + 1)(x - 3) $

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11. Verify if -2 and 3 are zeroes of $ 2x^3 - 3x^2 - 11x + 6 $. If yes, factorize.

Let $ p(x) = 2x^3 - 3x^2 - 11x + 6 $

Check $ x = -2 $:
$$
2(-8) - 3(4) - 11(-2) + 6 = -16 - 12 + 22 + 6 = (-28) + 28 = 0 ✔
$$

Check $ x = 3 $:
$$
2(27) - 3(9) - 11(3) + 6 = 54 - 27 - 33 + 6 = (54 - 27) = 27; (27 - 33) = -6; (-6 + 6) = 0 ✔
$$

So both are zeros → $ (x + 2) $ and $ (x - 3) $ are factors.

Multiply: $ (x + 2)(x - 3) = x^2 - x - 6 $

Now divide $ p(x) $ by this quadratic.

Use polynomial division or factor.

Let’s use factoring.

Since it's cubic, and we have two roots, find third.

Sum of roots: $ -(-3)/2 = 3/2 $

Roots: $ -2, 3, r $

Then $ -2 + 3 + r = 3/2 \Rightarrow 1 + r = 1.5 \Rightarrow r = 0.5 = 1/2 $

So $ x = 1/2 $ is a root → $ (2x - 1) $ is a factor.

Now verify:

Let’s do synthetic division with $ x = 1/2 $:

```
1/2 | 2 -3 -11 6
| 1 -1 -6
-------------------
2 -2 -12 0
```

Quotient: $ 2x^2 - 2x - 12 = 2(x^2 - x - 6) = 2(x - 3)(x + 2) $

So full factorization:
$$
(2x - 1)(x + 2)(x - 3)
$$

✔ Answer: Yes, both are zeros. Factorized form: $ (2x - 1)(x + 2)(x - 3) $

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12. Verify if 1 and -3 are zeroes of $ 3x^3 + 5x^2 - 11x + 3 $. If yes, factorize.

Let $ p(x) = 3x^3 + 5x^2 - 11x + 3 $

Check $ x = 1 $:
$$
3 + 5 - 11 + 3 = 11 - 11 = 0 ✔
$$

Check $ x = -3 $:
$$
3(-27) + 5(9) - 11(-3) + 3 = -81 + 45 + 33 + 3 = (-81 + 45) = -36; +33 = -3; +3 = 0 ✔
$$

So both are zeros → $ (x - 1) $ and $ (x + 3) $ are factors.

Multiply: $ (x - 1)(x + 3) = x^2 + 2x - 3 $

Now divide $ p(x) $ by this.

Use synthetic division first with $ x = 1 $:

```
1 | 3 5 -11 3
| 3 8 -3
-----------------
3 8 -3 0
```

Quotient: $ 3x^2 + 8x - 3 $

Now factor $ 3x^2 + 8x - 3 $

Find two numbers: product $ 3×(-3) = -9 $, sum 8 → $ 9 $ and $ -1 $

Split:
$$
3x^2 + 9x - x - 3 = 3x(x + 3) -1(x + 3) = (3x - 1)(x + 3)
$$

So full factorization:
$$
(x - 1)(3x - 1)(x + 3)
$$

✔ Answer: Yes, both are zeros. Factorized: $ (x - 1)(x + 3)(3x - 1) $

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## ✔ Final Answers Summary:

1. $ a = \pm 1 $
2. $ k = -3 $
3. $ k = -2 - \sqrt{2} $
4. $ (3x - 1)(2x + 3) $
5. $ 3xy(x + y) $
6. Shown via substitution
7. $ (x - 1)(3x + 1)^2 $
8. $ (x + 1)(x + 3)(x - 2) $
9. $ (x - 3)(x - 4)(x - 5) $
10. Factor Theorem stated; factorization: $ (x - 1)(x + 1)(x - 3) $
11. Yes; $ (2x - 1)(x + 2)(x - 3) $
12. Yes; $ (x - 1)(x + 3)(3x - 1) $

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