Questions on the use of special polynomial forms such as the difference of two squares, the trinomial perfect square and sum and difference of two cubes to factor other polynomials are presented. Questions are presented along with detailed Solutions and explanations. We present several special polynomial forms.
We can write:
\[ 16x^2 - 9y^2 = (4x)^2 - (3y)^2 \]Now that we have written the given polynomial as the difference of two squares, we apply the identity:
\[ a^2 - b^2 = (a - b)(a + b) \]Using this identity to factor the polynomial:
\[ 16x^2 - 9y^2 = (4x - 3y)(4x + 3y) \]a) \[ a^2 + 2ab + b^2 = (a + b)^2 \] b) \[ a^2 - 2ab + b^2 = (a - b)^2 \]
Note that the monomials in the given polynomial may be written as follows:
\[ 4x^2 = (2x)^2 \quad , \quad 20xy = 2(2x)(5y) \quad \text{and} \quad 25y^2 = (5y)^2 \]We now write the given polynomial as follows:
\[ 4x^2 + 20xy + 25y^2 = (2x)^2 + 2(2x)(5y) + (5y)^2 \]Use the identity \( a^2 + 2ab + b^2 = (a + b)^2 \) to write the polynomial as a square:
\[ 4x^2 + 20xy + 25y^2 = (2x)^2 + 2(2x)(5y) + (5y)^2 = (2x + 5y)^2 \]Note that the monomials making up the given polynomial can be rewritten as follows:
\[ 1 = 1^2 \] \[ -6x = -2 \cdot 3 \cdot x \] \[ 9x^2 = (3x)^2 \]The given polynomial may therefore be written as:
\[ 1 - 6x + 9x^2 = 1^2 - 2 \cdot 3x + (3x)^2 \]Using the identity:
\[ a^2 - 2ab + b^2 = (a - b)^2 \]we recognize the pattern and write the given polynomial as a perfect square:
\[ 1 - 6x + 9x^2 = (1 - 3x)^2 \]Note that the monomials in the given polynomial can be rewritten as follows:
\[ 8 = (2)^3 \quad \text{and} \quad 27x^3 = (3x)^3 \]The given polynomial may now be written using cube powers:
\[ 8 - 27x^3 = (2)^3 - (3x)^3 \]We apply the identity for the difference of cubes:
\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]Using this identity, we factor the polynomial as follows:
\[ 8 - 27x^3 = (2)^3 - (3x)^3 = (2 - 3x)\left((2)^2 + (2)(3x) + (3x)^2\right) \] \[ = (2 - 3x)(4 + 6x + 9x^2) \]Thus, the factored form of the polynomial \( 8 - 27x^3 \) is:
\[ (2 - 3x)(9x^2 + 6x + 4) \]The two monomials making up the given polynomial may be written as:
\[ 8y^3 = (2y)^3 \quad \text{and} \quad 1 = (1)^3 \]The polynomial to factor may now be written as:
\[ 8y^3 + 1 = (2y)^3 + (1)^3 \]Use the identity for the sum of cubes:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]Apply this formula to factor the given polynomial:
\[ 8y^3 + 1 = (2y)^3 + (1)^3 = (2y + 1)\left((2y)^2 - (2y)(1) + (1)^2\right) \]Simplify the expression:
\[ (2y + 1)(4y^2 - 2y + 1) \]Final factored form:
\[ 8y^3 + 1 = (2y + 1)(4y^2 - 2y + 1) \]a) \[ -25x^2 + 9 \] b) \[ 16y^4 - x^4 \] c) \[ 36y^2 - 60xy + 25x^2 \] d) \[ \frac{1}{2}x^2 + x + \frac{1}{2} \] e) \[ -y^3 - 64 \] f) \[ x^6 - 1 \]
If we let \( a = 5x \) and \( b = 3 \), the given polynomial may be written as:
\[ -25x^2 + 9 = -a^2 + b^2 \]Use the special polynomial identity \( a^2 - b^2 = (a - b)(a + b) \) to factor the given expression as follows:
\[ -25x^2 + 9 = -a^2 + b^2 = (-a + b)(a + b) \]Now substitute back \( a = 5x \) and \( b = 3 \):
\[ (-5x + 3)(5x + 3) \] b)The given polynomial has the form of the difference of two squares and may be written as:
\[ 16y^4 - x^4 = (4y^2)^2 - (x^2)^2 \]Using the special identity for the difference of squares:
\[ a^2 - b^2 = (a - b)(a + b) \]We factor the given polynomial as follows:
\[ 16y^4 - x^4 = (4y^2)^2 - (x^2)^2 = (4y^2 - x^2)(4y^2 + x^2) \]The expression \( 4y^2 + x^2 \) is a sum of squares and cannot be factored using real numbers. However, the term \( 4y^2 - x^2 \) is also a difference of squares and can be factored further. Therefore, the complete factorization of the polynomial is:
\[ 16y^4 - x^4 = (2y - x)(2y + x)(4y^2 + x^2) \] c)The given polynomial may be written as:
\[ 36y^2 - 60xy + 25x^2 = (6y)^2 - 2(6y)(5x) + (5x)^2 \]Use the special trinomial \( a^2 - 2ab + b^2 = (a - b)^2 \) to factor the given polynomial as follows:
\[ 36y^2 - 60xy + 25x^2 = (6y)^2 - 2(6y)(5x) + (5x)^2 = (6y - 5x)^2 \] d)Factor \( \frac{1}{2} \) out and rewrite the given polynomial as:
\[ \frac{1}{2} x^2 + x + \frac{1}{2} = \frac{1}{2} x^2 + 2 \cdot \frac{1}{2} x + \frac{1}{2} = \frac{1}{2} \left( x^2 + 2x + 1 \right) \]Use the special trinomial identity \( a^2 + 2ab + b^2 = (a + b)^2 \) to factor \( x^2 + 2x + 1 = x^2 + 2(x)(1) + 1^2 \), and the given polynomial as follows:
\[ \frac{1}{2} x^2 + x + \frac{1}{2} = \frac{1}{2} \left( x^2 + 2x + 1 \right) = \frac{1}{2} (x + 1)^2 \] e)Factor out \( -1 \) and rewrite the given polynomial as:
\[ - y^3 - 64 = - (y^3 + 64) = - (y^3 + 4^3) \]Use the identity \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \) to factor the given polynomial as follows:
\[ - y^3 - 64 = - (y^3 + 64) = - (y^3 + 4^3) \] \[ = -(y + 4)(y^2 - y(4) + 4^2) = -(y + 4)(y^2 - 4y + 16) \] f)Let us write the given polynomial as the difference of two squares as follows:
\[ x^6 - 1 = (x^3)^2 - (1)^2 \]Use the special difference of squares polynomial \( a^2 - b^2 = (a - b)(a + b) \) and factor the given polynomial as follows:
\[ x^6 - 1 = (x^3)^2 - (1)^2 = (x^3 - 1)(x^3 + 1) \]In the above, we have the product of the sum and difference of two cubes. Hence:
\[ x^6 - 1 = (x^3)^2 - (1)^2 = (x^3 - 1)(x^3 + 1) \] \[ = (x - 1)(x^2 + x + 1)(x + 1)(x^2 - x + 1) \]