Factorize the following questions?

Question

1) 4x^4 + 7x^2 - 9
2) 5x^3 +14x^2 - 53x - 10


Dont simply write the answers down, show the stpes and the logic for the 1st question , I tried to think for both the question for a long time but it just isnt cliking the rest were moder and easy and tough but I could do it correctly, Thanks for the help.

I believe it can be facotrized by factor method , I tried splittting the midle term method but didnt work, please tell me how to factorize by factor method

Answer 1

1) 4x^4 + 7x^2 - 9
4(x^4 + (7/4)x^2 - 9/4)
4(x^4 + (7/4)x^2 + (7/8)^2 - (7/8)^2 - 9/4)
4(x^2 + 7/8)^2 - 49/64 - 144/64)
4(x^2 + 7/8)^2 - 193/64)
4(x^2 + 7/8)^2 - ((1/16)√193)^2)
4(x^2 + 7/8 - (1/16)√193)(x^2 + 7/8 + (1/16)√193)

Answer 2

2) Regarding the second problem if the expression is 5x^3+14x^2-53x+10 instead of as given by yoy,it can be factorised.
5x^3+14x^2-53x+10
Applying the factor theorem which states that if an expression containing x,becomes zero when x is replaced by a,then x-a is a factor of the expression
If we put x=2,the value of the expression is
5*8+14*4-53*2+10
=40+56-106+10=0
Therefore x-2 is a factor of the polynomial
5x^3+14x^2-53x+10
=5x^2(x-2)+24x(x-2)-5(x-2)
=(x-2)(5x^2+24x-5)
=(x-2)(5x^2+25x-x-5)
=(x-2){5x(x+5)-1(x+5)}
=(x-2)(x+5)(5x-1)

If the question is from your text-book,please check the answer and if the same is tallying,then correct the question.Otherwise please ignore my answer.Thank you

Answer 3

both of them are not factorable.
the complication of the first question is the 4 infront of x^4
we know we will brake it into 2 ( ) ( ) with x^2 in the begining of both so thats the first step
(?x^2 +- ? ) ( ?x^2 +- ?) +- means plus or minus
the first coefeicinets infront of x^2 has to be factors of 4 that is infront of x^4. the only two ways we can get 4 by multiplication is 4x1 and 2x2 we have to try both of them
so its either (2x^2 +-?) (2x^2 +- ?) or (4x^2 +-?)(1x^2 +-?)
so now we have to look at 9 and think how can we factor that and the obly ways ti do that is
9x1 and 3x3
when we plug in them into each we get
(2x^2+-9)(2x^+-1) or (4x^2+-9)(x^2+-1) or (4x^2+-1)(x^2+-9)
(2x^2+-3)(2x^2+-3) or (4x^2+-3)(x^2+-3)
neither of these will work. so its not factorable.
the problem with the second one is #53. its a prime number and neither of the other coeficients will go into it. so it cant be factored.

Answer 4

ok you want to locate the worry-unfastened factors in the two factors; ie, selection one, the two factors have a p²and a q², so those 2 go exterior the bracket: p²q²() then interior the bracket is going despite you will possibly be able to desire to multiply those by utilising to get the unique words; ie, for the 1st time, you want yet another q, and for the 2nd, yet another p, making the factorised variety: p²q²(q+p). the 2nd is comparable. for the 0.33, you will possibly be able to desire to multiply out the brackets earlier factorising the whole element (keep in mind (x-one million)^2=(x-one million)(x-one million)!) desire that helps.