Another type of binomial expression that can be factorised very nicely are perfect squares. We've already come across some of these and know how to expand them:
$\left(a+b\right)^2=a^2+2ab+b^2$(a+b)2=a2+2ab+b2
$\left(a-b\right)^2=a^2-2ab+b^2$(a−b)2=a2−2ab+b2
Notice a pattern in the expansion of $\left(a+b\right)^2$(a+b)2:
$a^2$a2 | $+$+ | $2ab$2ab | $+$+ | $b^2$b2 |
---|---|---|---|---|
$\uparrow$↑ | $\uparrow$↑ | $\uparrow$↑ | ||
Square of the first term | Double the product of the two terms | Square of the second term |
So if we have an expanded expression that fits this pattern, we can go backwards and factorise it.
That is:
$a^2+2ab+b^2=\left(a+b\right)^2$a2+2ab+b2=(a+b)2
There is a very similar result with the differences of two squares, where we used the result of expansion to go backwards and factorise.
Which of the following are perfect square expansions?
A) $x^2+6x+9$x2+6x+9
Solution:
We can write $x^2+6x+9$x2+6x+9 as $\left(x\right)^2+2\left(3x\right)+\left(3\right)^2$(x)2+2(3x)+(3)2
We have the square of $x$x, the square of $3$3, and double the product of $x$x and $3$3. So this is a perfect square expression that can be factorised.
B) $x^2-4x+16$x2−4x+16
Solution:
We can write $x^2-4x+16$x2−4x+16 as $\left(x\right)^2-4x+\left(-4\right)^2$(x)2−4x+(−4)2
We have the square of $x$x and the square of $\left(-4\right)$(−4). But the other term, $-4x$−4x is not double the product of $x$x and $\left(-4\right)$(−4). So this is not a perfect square expression.
C) $x^2-18x+81$x2−18x+81
Solution:
We can write $x^2-18x+81$x2−18x+81 as $\left(x\right)^2+2\left(-9x\right)+\left(-9\right)^2$(x)2+2(−9x)+(−9)2
We have the square of $x$x, the square of $\left(-9\right)$(−9), and double the product of $x$x and $\left(-9\right)$(−9). So this is a perfect square expression that can be factorised.
D) $9+12x+4x^2$9+12x+4x2
Solution:
We can write $9+12x+4x^2$9+12x+4x2 as $\left(3\right)^2+2\left(3\times2x\right)+\left(2x\right)^2$(3)2+2(3×2x)+(2x)2
We have the square of $3$3, the square of $2x$2x, and double the product of $3$3 and $2x$2x. So this is a perfect square expression that can be factorised.
By expanding $\left(b+7\right)^2$(b+7)2, we want to determine the factorisation for $b^2+14b+49$b2+14b+49.
First expand $\left(b+7\right)^2$(b+7)2.
Hence, factorise $b^2+14b+49$b2+14b+49.
Factorise the expression $36y^2-12yz+z^2$36y2−12yz+z2
Think about how to attain $a$a and $b$b, and how to check if this really is a perfect square expression.
Do
$\sqrt{36y^2}$√36y2 | $=$= | $\sqrt{36}\sqrt{y^2}$√36√y2 | |
$=$= | $6y$6y | ||
$\sqrt{z^2}$√z2 | $=$= | $z$z | |
Therefore | $a$a is $6y$6y and our $b$b is $z$z. | ||
$2\times a\times b$2×a×b | $=$= | $2\times6y\times z$2×6y×z | |
$=$= | $12yz$12yz | ||
which is the same as the middle term so this is a perfect square expression | |||
So | |||
$36y^2-12yz+z^2$36y2−12yz+z2 | $=$= | $\left(6y-z\right)^2$(6y−z)2 |
Factorise $x^2-12x+36$x2−12x+36.
Factorise the following expression completely: $3p^2+12p+12$3p2+12p+12
Think about how to make this a perfect square expression by first factorising all the coefficients.
Do: We can see here that $3p^2$3p2 and $12$12 are not perfect squares! So how can we find $a$a and $b$b?
If we look at this another way we can see that the coefficients of all three terms have an HCF of $3$3, so let's factorise that out first.
$3p^2+12p+12=3\left(p^2+4p+4\right)$3p2+12p+12=3(p2+4p+4)
Now the expression in the brackets is a perfect square expression where $\sqrt{p^2}=p$√p2=p and $\sqrt{4}=2$√4=2.
Therefore $3\left(p^2+4p+4\right)=3\left(p+2\right)^2$3(p2+4p+4)=3(p+2)2
Always see if you can factorise expressions using the most basic method of $AB+AC=A\left(B+C\right)$AB+AC=A(B+C) first if you can. It'll make things a lot easier later on!
Factorise $-18b^2-12ab-2a^2$−18b2−12ab−2a2.
Form and solve linear equations and inequations, quadratic and simple exponential equations, and simultaneous equations with two unknowns
Apply algebraic procedures in solving problems