How do you factor the trinomial $18 a^{2} + 48 a + 32$ $18a^2+48a+32$ ?

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Answer 1 · 2015-11-18T14:24:55

$2 {(3 a + 4)}^{2}$ $2(3a+4)^2$

$18 a^{2} + 48 a + 32$ $18a^2 + 48a + 32$

$\frac{18 a^{2}}{2} + \frac{48 a}{2} + \frac{32}{2}$ $(18a^2)/2 + (48a)/2 + 32/2$

$9 a^{2} + 24 a + 16$ $9a^2 + 24a + 16$

16(last term) x 9(co-efficient of $a^{2}$ $a^2$ ) = 144

Think of two numbers that
$\Rightarrow$ $rArr$ When you multiply them = 144
$\Rightarrow$ $rArr$ When you add them = 24 (co-efficient of $a$ $a$ )
12 and 12
$\Rightarrow$ $rArr$ 12 x 12 = 144
$\Rightarrow$ $rArr$ 12 + 12 = 24
Replace the middle number with 12 and 12
$9 a^{2} + 12 a + 12 a + 16$ $9a^2 + 12a + 12a + 16$
Split the expression
$9 a^{2} + 12 a - - - - - ∣ - - - - - + 12 a + 16$ $9a^2 + 12a----- | ----- + 12a + 16$
Factorise the two
$3 a (3 a + 4) + 4 (3 a + 4)$ $3a(3a + 4) + 4(3a +4)$
Common factors so group
$(3 a + 4) (3 a + 4)$ $(3a + 4)(3a +4)$

$\Rightarrow$ $rArr$ ${(3 a + 4)}^{2}$ $(3a+4)^2$

How do you factor the trinomial 18a2+48a+3218a^2+48a+32?