How do you solve $3 x^{2} + 17 x + 20 = 0$ $3x^2+17x+20=0$ ?

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Answer 1 · 2018-05-05T02:45:41

$x = \frac{- 5}{3}$ $x = (-5)/3$ or $x = - 4$ $x=-4$

$3 x^{2} + 17 x + 20 = 0$ $3x^2 +17x+20=0$

Can be factored by finding the factors for $a \cdot c$ $a*c$

$3 \cdot 20 = 60$ $3*20=60$

The factors of 60 are
$1 \cdot 60$ $1*60$
$2 \cdot 30$ $2*30$
$3 \cdot 20$ $3*20$
$4 \cdot 15$ $4*15$
$5 \cdot 12$ $5*12$
$6 \cdot 10$ $6*10$

The factor pattern that adds up to the middle term is $5 + 12 = 17$ $5+12 = 17$

Set the Polynomial up as

$3 x^{2} + 12 x + 5 x + 20$ $3x^2 +12x +5x +20$

Now factor by grouping

$[3 x^{2} + 12 x] [+ 5 x + 20]$ $[3x^2 +12x][ +5x +20]$

$3 x (x + 4) + 5 (x + 4)$ $3x(x+4)+5(x+4)$

So the common factor is $(x + 4)$ $(x+4)$

$(3 x + 5) (x + 4) = 0$ $(3x+5)(x+4)=0$

Set both factors equal to zero

$3 x + 5 = 0$ $3x+5 = 0$

$x = - \frac{5}{3}$ $x=-5/3$

$x + 4 = 0$ $x+4=0$

$x = - 4$ $x=-4$

How do you solve 3x2+17x+20=03x^2+17x+20=0?