How do you factor $2 m^{2} + 19 m + 9$ $2m ^ { 2} + 19m + 9$ ?

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Answer 1 · 2018-03-15T02:21:21

$m = - \frac{1}{2} and m = - 9$ $m= -1/2 and m=-9$

$2 m^{2} + 19 m + 9$ $2m^2+19m+9$

multiply the first and last terms to for the number that you will find the factors of
$2 m^{2} \cdot 9 = 18 m^{2}$ $2m^2*9=18m^2$

find the numbers that multiply to 18 and add to 19
$1 \cdot 18 = 18; 1 + 18 = 19$ $1*18=18; 1+18=19$

put those numbers in place of the multiplied number you found and drop down the first and last term
$2 m^{2} + 18 m + (1) m + 9$ $2m^2+18m+(1)m + 9$

group the terms into two groups of two using parentheses
$(2 m^{2} + 18 m) + (m + 9)$ $(2m^2+18m)+(m+9)$

factor the groups, the parentheses should both be the same
$2 m (m + 9) + 1 (m + 9)$ $2m(m+9)+1(m+9)$

the values in the parentheses are one of your answers, the ones outside are your other answer.
$m = (2 m + 1); m = (m + 9)$ $m=(2m+1) ; m=(m+9)$

set them equal to zero to find the value
$0 = 2 m + 1; 0 = m + 9$ $0=2m+1 ;0=m+9$

$- \frac{1}{2} = m; - 9 = m$ $-1/2=m; -9=m$

How do you factor 2m2+19m+92m ^ { 2} + 19m + 9?