How do you factor #25r ^ { 2} - 10r - 35#?

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Answer 1 · 2017-03-26T20:26:16

# 5 * ( 5r - 7) * ( r + 1) #

You solve it for zero using the discriminant

# 25r^2 - 10r - 35 = 0 #

Δ= #10^2 - 4 * 25 * (-35) # = # 100 + 3500 # = 3600

then r1= # ( - (-10) + sqrt3600) / ( 2 * 25) # = # (10 +60 ) / 50 # = #7/5#

and r2# ( - (-10) - sqrt3600) / ( 2 * 25) # = # (10 - 60) / 50 # = -1

so you use the formula

# a * ( r - r1 ) * ( r - r2) # and

your answer is ....
# 25 * ( r - 7/5 ) * (r - (-1) ) #
or further simplified :

# 5 * ( 5r - 7) * ( r + 1) #

Answer 2 · 2017-03-26T20:41:21

#25r^2-10r-35#
#=5(5r+7)(r-1)#

#=25(r+7/5)(r-1)#

#25r^2-10r-35#

#=5[5r^2color(blue)(-2r)-7]#

#=5[5r^2color(blue)(-5r+7r)-7]#

#=5[color(green)(5r)color(red)((r-1))+color(green)(7)color(red)((r-1))]#

#=5color(green)((5r+7))color(red)((r-1))#

#=25(r+7/5)(r-1)#