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OpenStudy (ajprincess):
Pls tell me two factors of -10 so that their sum is -3.
OpenStudy (anonymous):
5 and 2 one negatve one positive
OpenStudy (anonymous):
-5 and 2
OpenStudy (ajprincess):
yup good:)
Nw -3x can be written as -5x+2x. Agree?
OpenStudy (anonymous):
yes
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OpenStudy (ajprincess):
good:)
\(x^2-3x-10=0\)
\(x^2-5x+2x-10=0\)
what do u get when u factorise \(x^2-5x\)?
OpenStudy (anonymous):
ughhhhhh im stuck lol
OpenStudy (ajprincess):
\(x^2-5x=x*x-5*x\)
What is common to both \(x^2\) and \(5x\)?
OpenStudy (anonymous):
x
OpenStudy (ajprincess):
good:) nw when u take it out what remains
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OpenStudy (anonymous):
x-5
OpenStudy (anonymous):
aj?
OpenStudy (ajprincess):
ya. right:)
So
\(x^2-5x=x(x-5)\)
Nw can u factorise
2x-10
OpenStudy (anonymous):
x-5
OpenStudy (ajprincess):
well
2x-10=2(x-5)
Nw can u factorise
x(x-5)+2(x-5)
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OpenStudy (ajprincess):
@kittenbalm
OpenStudy (anonymous):
ugh how do i foil?
OpenStudy (ajprincess):
treat (x-5) as one term. for simplicity let us assume (x-5)=y.
xy+2y. can u factorise this?
OpenStudy (anonymous):
yes you can
OpenStudy (ajprincess):
I was asking whether u can? I can
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OpenStudy (anonymous):
oh no
OpenStudy (ajprincess):
xy+2y
What is common to both xy and 2y?
OpenStudy (anonymous):
y
OpenStudy (ajprincess):
good:)
So what remains when u take it out?
OpenStudy (anonymous):
x+2
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OpenStudy (ajprincess):
right:)
xy+2y=y(x+2)
our y=x-5.
so
x(x-5)+2(x-5)=(x-5)(x+2)
\(x^2-3x-10=0\)
\(x^2-5x+2x-10=0\)
\(x(x-5)+2(x-5)=0\)
\((x-5)(x+2)=0\)
x-5=0 or x+2=0
Hope u can find the two values of x nw:)