Factorisation gets really interesting when we start to deal with more complicated algebraic polynomials, which are just big chains of algebraic expressions. One of these is called a binomial (you might have encountered them already), which is any polynomial with just two terms, usually not alike. The 'bi' in the word gave it away didn't it! For example $x+y$x+y, $4-w$4−w, $-z+55$−z+55 are all binomials.
Really interesting patterns appear when we multiply two binomials together, and you here we're going to look at one of the form $\left(a+b\right)\left(a-b\right)$(a+b)(a−b), which you might remember as being called difference of two squares.
So now we know that:
$\left(a+b\right)\left(a-b\right)=a^2-b^2$(a+b)(a−b)=a2−b2
Be careful that in this example where $a$a is the same sign in both brackets the answer is always $a^2-b^2$a2−b2 and NOT $b^2-a^2$b2−a2! We already know how to use the above property to expand the bracketed expression on the left, but now we can also use it to factorise the one on the right, since factorisation is the opposite of expansion.
Answer the following.
Expand $\left(x+5\right)\left(x-5\right)$(x+5)(x−5).
Hence factorise $x^2-25$x2−25.
Factorise $121m^2-64$121m2−64.
Factorise $3t^2-12$3t2−12.
Factorise the expression $16x^2-81y^2$16x2−81y2.
Think about how we should go from $a^2$a2 and $b^2$b2 terms in the property above to $a$a's and $b$b's by finding square roots.
Do
$\sqrt{16x^2}$√16x2 | $=$= | $\sqrt{16}\sqrt{x^2}$√16√x2 |
$=$= | $4x$4x | |
$\sqrt{81y^2}$√81y2 | $=$= | $\sqrt{81}\sqrt{y^2}$√81√y2 |
$=$= | $9y$9y | |
Therefore | ||
$\left(4x\right)^2$(4x)2 | $=$= | $16x^2$16x2 |
and | ||
$\left(9y\right)^2$(9y)2 | $=$= | $81y^2$81y2 |
So | ||
$16x^2-81y^2$16x2−81y2 | $=$= | $\left(4x+9y\right)\left(4x-9y\right)$(4x+9y)(4x−9y) |
Form and solve linear equations and inequations, quadratic and simple exponential equations, and simultaneous equations with two unknowns
Apply algebraic procedures in solving problems