hcrossman29
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Factor 3y2 + 29y − 10.
3y2 + 29y – 10 = (________ − 1)( _________ + 10)
I want to see the work too because i am a bit confused.

Respuesta :

quadratic trinomial can be represent as product
ax² + bx + c =a(x - x1)(x - x2), where x1 and x2 are roots of quadratic equation ax² + bx + c=0,
so to solve this problem we need to find roots of 
3y²+29y-10=0

y=(-b+/-√(b²-4ac))/2a
y=(-29+/-√(29²-4*3*(-10)))/(2*3)
y=(-29+/-√(961)/(6)
y=(-29+/-31/(6)
y1=-60/6=-10
y2=2/6=1/3

3(y-1/3)(y+10)=(3y-1)(y+10)
check:   (3y-1)(y+10)=3y²-y+30y-10=3y²+29y-10

Answer:

Factors (3y -1) ( y + 10 ).

Step-by-step explanation:

Given :  3y² + 29y − 10.

To find : Fill  (________ − 1)( _________ + 10)

Solution : We have given

3y² + 29y − 10.

On factoring

3y² + 30y - y − 10.  

Taking common 3y form first two terms and -1 from last two terms.

3y ( y + 10 ) -1( y + 10 ) .

On grouping

(3y -1) ( y + 10 ).

Therefore, Factors (3y -1) ( y + 10 ).

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