2.3-Factorization of Polynomials

2.3-Factorization of Polynomials Important Formulae

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Grade 9 → Math → Polynomials → 2.3-Factorization of Polynomials

After successful completion of this topic, you should be able to:

  • Factorize a polynomial.

Factorization of polynomials is the process of breaking down a polynomial into simpler factors that, when multiplied together, give the original polynomial. It is an essential skill in algebra, allowing us to solve equations and understand polynomial behavior.

Types of Factorization:
1. Common Factor Method: The first step in factorization is to find the greatest common factor (GCF) of the terms in the polynomial. For example, consider the polynomial $2x^3 + 4x^2$. The GCF is $2x^2$, so we factor it out: $$ 2x^3 + 4x^2 = 2x^2(x + 2). $$ 2. Grouping Method: When a polynomial has four or more terms, we can use the grouping method. We group terms in pairs and factor out the GCF from each pair. For example: $$ x^3 + 2x^2 + x + 2 = (x^3 + 2x^2) + (x + 2) = x^2(x + 2) + 1(x + 2) = (x + 2)(x^2 + 1). $$ 3. Quadratic Trinomials: A polynomial of the form $ax^2 + bx + c$ can often be factored into the product of two binomials. For example: $$ x^2 + 5x + 6 = (x + 2)(x + 3). $$ We look for two numbers that multiply to $c$ (6) and add to $b$ (5).
4. Difference of Squares: Polynomials of the form $a^2 - b^2$ can be factored using the identity: $$ a^2 - b^2 = (a - b)(a + b). $$ 5. Perfect Square Trinomials: A perfect square trinomial can be factored into: $$ a^2 + 2ab + b^2 = (a + b)^2, $$ and $$ a^2 - 2ab + b^2 = (a - b)^2. $$ 6. Cubic Polynomials: Cubic polynomials can often be factored by finding one root and using synthetic division. For instance, for the polynomial $x^3 - 6x^2 + 11x - 6$, we can check possible rational roots using the Rational Root Theorem. After finding a root (e.g., $x = 1$), we factor as follows: $$ x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6) = (x - 1)(x - 2)(x - 3). $$
Special Cases:
- Sum of Cubes: The identity for the sum of cubes is: $$ a^3 + b^3 = (a + b)(a^2 - ab + b^2). $$ - Difference of Cubes: The identity for the difference of cubes is: $$ a^3 - b^3 = (a - b)(a^2 + ab + b^2). $$


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Determine which of the following polynomials has (x + 1) a factor :

(i) x$^3$ +x$^2$ +x + 1
(ii) x$^4$ +x$^3$ +x$^2$ +x + 1
(iii)x$^4$+3x$^3$+3x$^2$ +x +1
(iv)x$^3$–x$^2$–(2+ $\sqrt{2}$)x+ 2

Solution:

Determining Factors of Polynomials

To check if $(x + 1)$ is a factor, we substitute $x = -1$ into each polynomial:

(i) $p(-1) = (-1)^3 + (-1)^2 + (-1) + 1 = 0$ (Factor)
(ii) $p(-1) = (-1)^4 + (-1)^3 + (-1)^2 + (-1) + 1 = 0$ (Factor)
(iii) $p(-1) = (-1)^4 + 3(-1)^3 + 3(-1)^2 + (-1) + 1 = 0$ (Factor)
(iv) $p(-1) = (-1)^3 - (-1)^2 - (2 + \sqrt{2})(-1) + 2 = 1 + 2 + \sqrt{2} \neq 0$ (Not a Factor)

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Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:

(i) p(x)=$2x^3 +x^2 –2x–1$, g(x)=x+1
(ii) p(x)=$x^3 +3x^2 +3x+1$, g(x)=x+2
(iii) p(x)=$x^3 –4x^2 +x+6$, g(x)=x–3

Solution:
(i) Yes
(ii) No
(iii) Yes

Using the Factor Theorem

(i) For $p(x)=2x^3 +x^2 –2x–1$ and $g(x)=x+1$, find $p(-1)$. If $p(-1)=0$, then $g(x)$ is a factor.

(ii) For $p(x)=x^3 +3x^2 +3x+1$ and $g(x)=x+2$, find $p(-2)$. If $p(-2)\neq 0$, then $g(x)$ is NOT a factor.

(iii) For $p(x)=x^3 –4x^2 +x+6$ and $g(x)=x–3$, find $p(3)$. If $p(3)=0$, then $g(x)$ is a factor.

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Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:

(i) p(x)= $x^2 + x + k$
(ii) p(x)=$2x^2 +kx+ \sqrt{2}$
(iii) p(x)=$kx^2 – \sqrt{2}x+1$
(iv) p(x)=$kx^2 –3x+k$

Solution:

Finding Value of k

For $x - 1$ to be a factor, $p(1) = 0$: (i) $p(1) = 1^2 + 1 + k = 0 \implies k = -2$.
(ii) $p(1) = 2(1^2) + k(1) + \sqrt{2} = 0 \implies k = -2 - \sqrt{2}$.
(iii) $p(1) = k(1^2) - \sqrt{2}(1) + 1 = 0 \implies k = \sqrt{2} - 1$.
(iv) $p(1) = k(1^2) - 3(1) + k = 0 \implies 2k = 3 \implies k = \frac{3}{2}$.

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Factorise:

(i) $12x^2 –7x+1$
(ii) $2x^2 +7x+3$
(iii) $6x^2 +5x–6$
(iv) $3x^2–x–4$

Solution:

Factorization of Polynomials

(i) $12x^2 - 7x + 1 = (3x - 1)(4x - 1)$.
(ii) $2x^2 + 7x + 3 = (2x + 1)(x + 3)$.
(iii) $6x^2 + 5x - 6 = (2x - 1)(3x + 6)$.
(iv) $3x^2 - x - 4 = (3x - 4)(x + 1)$.

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Factorise:

(i) $x^3 –2x^2 –x+2$
(ii) $x^3 –3x^2 –9x–5$
(iii) $x^3 +13x^2+32x+20$
(iv) $2y^3 +y^2 –2y–1$

Solution:

Factorization of Polynomials

(i) $x^3 - 2x^2 - x + 2 = (x - 2)(x^2 + 1)$.
(ii) $x^3 - 3x^2 - 9x - 5 = (x + 1)(x^2 - 4x - 5) = (x + 1)(x - 5)(x + 1)$.
(iii) $x^3 + 13x^2 + 32x + 20 = (x + 1)(x^2 + 12x + 20) = (x + 1)(x + 2)(x + 10)$.
(iv) $2y^3 + y^2 - 2y - 1 = (2y + 1)(y^2 - 1) = (2y + 1)(y - 1)(y + 1)$.

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