How do you factor the trinomial #7y^2- 50y + 7#?

2 Answers
George C.
Feb 29, 2016

#7y^2-50y+7 = (7y-1)(y-7)#

Explanation:

Let #f(y) = 7y^2-50y+7#

By the rational root theorem, any rational zeros of #f(y)# must be expressible in the form #p/q# for some integers #p# and #q# where #p# is a divisor of the constant term #7# and #q# is a divisor of the coefficient #7# of the leading term.

That means that the only possible rational zeros are:

#+-1/7, +-1, +-7#

In addition, since the coefficients are symmetric, if #y=r# is a zero then so is #y=1/r#.

Also we find that #f(-y) = 7y^2+50y+7# has no changes of sign in its coefficients, so #f(y)# has no negative zeros.

Also the sum of the coefficients is non-zero. That is #7-50+7 = -36 != 0#, so #y=1# is not a zero.

Hence if the zeros are rational, then they are #1/7# and #7#.

Try:

#(7y-1)(y-7) = 7y^2-50y+7#

Yes!

George C.
Feb 29, 2016

Use an AC Method to find:

#7y^2-50y+7=(7y-1)(y-7)#

Explanation:

Given #7y^2-50y+7# find a pair of factors of #AC=7*7=49# with sum #B=50#.

The pair #49, 1# works:

#49xx1 = 49#

#49+1 = 50#

Use this pair to split the middle term and factor by grouping:

#7y^2-50y+7#

#=7y^2-49y-y+7#

#=(7y^2-49y)-(y-7)#

#=7y(y-7)-1(y-7)#

#=(7y-1)(y-7)#

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