How do you factor #7x^2 – 35x + 28#?

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Answer 1 · 2018-07-27T11:20:39

#7x^2-35x+28=7(x-4)(x-1)#

First, take out the common factor which is #7#

#7x^2-35x+28#

#=7(x^2-5x+4)#

#x=(5+-sqrt(25-4(1)(4)))/(2(1))#

#x=(5+-sqrt9)/2#

#x=(5+-3)/2#

#x=(5-3)/2# or #x=(5+3)/2#

#x=1# or #x=4#

#=7(x^2-5x+4)#

#=7(x-4)(x-1)#

Answer 2 · 2018-07-27T14:04:22

#7(x-1)(x-4)#

#"take out a "color(blue)"common factor "7#

#=7(x^2-5x+4)#

#"the factors of "+4" which sum to "-5#

#"are "-1" and "-4#

#=7(x-1)(x-4)#

Answer 3 · 2018-07-28T00:24:47

#7(x-1)(x-4)#

Since all terms are divisible by #7#, we can factor this out to get

#7(x^2-5x+4)#

What we have in parenthesis can be factored with a little thought experiment:

What two numbers sum up to #-5# and have a product of #4#?

After some trial and error, we arrive at #-1# and #-4#. This means we can factor this as

#7(x-1)(x-4)#

Hope this helps!