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Factorize each of the following quadratic polynomials by using the method of completing the square:
(i) $y^2-7y+12$
(ii) $z^2-4z-12$
Given:
The given quadratic polynomials are:
(i) $y^2-7y+12$
(ii) $z^2-4z-12$
To do:
We have to factorize the given quadratic polynomials.
Solution:
Factorizing algebraic expressions:
Factorizing an algebraic expression implies writing the expression as a product of two or more factors. Factorization is the reverse of distribution.
An algebraic expression is factored completely when it is written as a product of prime factors.
Completing the square is a method that is used to write a quadratic expression in a way such that it contains the perfect square.
(i) The given expression is $y^2-7y+12$.
Here,
The coefficient of $y^2$ is $1$
The coefficient of $y$ is $-7$
The constant term is $12$
Coefficient of $y^2$ is $1$. So, we can factorize the given expression by adding and subtracting square of half of coefficient of $y$.
Therefore,
$y^2-7y+12=y^2-7y+12+(\frac{7}{2})^2-(\frac{7}{2})^2$ [Since $\frac{1}{2}\times7=\frac{7}{2}$]
$y^2-7y+12=y^2-2(y)(\frac{7}{2})+(\frac{7}{2})^2+12-(\frac{7}{2})^2$
$y^2-7y+12=(y-\frac{7}{2})^2+12-\frac{49}{4}$ (Completing the square)
$y^2-7y+12=(y-\frac{7}{2})^2+\frac{12\times4-49}{4}$
$y^2-7y+12=(y-\frac{7}{2})^2+\frac{48-49}{4}$
$y^2-7y+12=(y-\frac{7}{2})^2-\frac{1}{4}$
Now,
$(y-\frac{7}{2})^2-\frac{1}{4}$ can be written as,
$(y-\frac{7}{2})^2-\frac{1}{4}=(y-\frac{7}{2})^2-(\frac{1}{2})^2$ [Since $\frac{1}{4}=(\frac{1}{2})^2$]
Using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize the given expression as,
$(y-\frac{7}{2})^2-\frac{1}{4}=(y-\frac{7}{2})^2-(\frac{1}{2})^2$
$(y-\frac{7}{2})^2-\frac{1}{4}=(y-\frac{7}{2}+\frac{1}{2})(y-\frac{7}{2}-\frac{1}{2})$
$(y-\frac{7}{2})^2-\frac{1}{4}=(y+\frac{-7+1}{2})(y-\frac{7+1}{2})$
$(y-\frac{7}{2})^2-\frac{1}{4}=(y+\frac{-6}{2})(y-\frac{8}{2})$
$(y-\frac{7}{2})^2-\frac{1}{4}=(y-3)(y-4)$
Hence, the given expression can be factorized as $(y-4)(y-3)$.
(ii) The given expression is $z^2-4z-12$.
Here,
The coefficient of $z^2$ is $1$
The coefficient of $z$ is $-4$
The constant term is $-12$
Coefficient of $z^2$ is $1$. So, we can factorize the given expression by adding and subtracting square of half of coefficient of $z$.
Therefore,
$z^2-4z-12=z^2-4z-12+2^2-2^2$ [Since $\frac{1}{2}\times4=2$]
$z^2-4z-12=z^2-4z+2^2-12-4$
$z^2-4z-12=z^2-2(z)(2)+2^2-16$
$z^2-4z-12=(z-2)^2-16$ (Completing the square)
Now,
$(z-2)^2-16$ can be written as,
$(z-2)^2-16=(z-2)^2-4^2$ [Since $16=4^2$]
Using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize the given expression as,
$(z-2)^2-16=(z-2)^2-4^2$
$(z-2)^2-16=(z-2+4)(z-2-4)$
$(z-2)^2-16=(z+2)(z-6)$
Hence, the given expression can be factorized as $(z-6)(z+2)$.
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