Write f(x) = x^4-10x^2-4x+8 and observe . 1. f(x) is +ve for x large positive or negative and for x=0, 2. f(x) is -ve for x = +2 or -2, so there are four real roots. Before assuming quadratic product you should eliminate (x-a)*(x^3+bx^2+cx+d)... and only then look for quadratic. a*d cannot have a or d odd since the b,c coefficients -10, -4 are both even, so only a=(-4,-2,+2,+4) and f(x) is non-zero for any of those. So there is no rational root a. Then we look for (x^2-ax+b)(x^2+ax+c) with a>0, since only this form gives zero x^3 coefficient x^2 coefficient is -10, so b+c-a^2 = -10 and a(b-c) = -4 Look for b and c +-2,+-4 such that bc=8, with b, c having the same sign (-2,-4), (-4,-2), (+2,+4), (+4,+2) Note that 16 - (4+2) = 10 = 4 +(4+2), and a=4 or 2 c>b since the x^2 coefficient (-ac+ab) = a*(b-c) =-10 which gives us (b,c) = (2,4) or (b,c) = (-4,-2) (b+c-a^2) = -10, a(b-c) = -4 so a!=4, a=2, b=-4, c=-2 f(x) = (x^2 - 2x - 4)*(x^2 + 2x - 2) Two quadratics each with real solutions since "b^2-4ac > 0" in both cases, but I won't bother calculating the four roots, that's trivial So with simple equations we deduce a,b,c without all the long winded unexplained mumbo-jumbo of forming two square expressions. There is nothing intuitive or formal in your method. The approach above is rigorous.
Eu sou professor de matemática de formação aqui no Brasil, mas não temos o costume de resolver equações com esses artifícios, portanto temos dificuldades com esse tipo de questão. (I am a mathematics teacher by training here in Brazil, but we are not used to solving equations with these devices, so we have difficulties with this type of question.)
Analisando com mais cuidado, percebi que o coeficiente do termo de segundo grau deve ser par, e deve ser desmembrado em duas parcelas, tais que uma seja par maior que 1 e a outra seja um quadrado perfeito. (Analyzing more carefully, I realized that the coefficient of the second degree term must be even, and must be divided into two parts, such that one is even greater than 1 and the other is a perfect square.)
Write f(x) = x^4-10x^2-4x+8 and observe .
1. f(x) is +ve for x large positive or negative and for x=0,
2. f(x) is -ve for x = +2 or -2,
so there are four real roots.
Before assuming quadratic product you should eliminate (x-a)*(x^3+bx^2+cx+d)... and only then look for quadratic.
a*d cannot have a or d odd since the b,c coefficients -10, -4 are both even, so only a=(-4,-2,+2,+4)
and f(x) is non-zero for any of those. So there is no rational root a.
Then we look for (x^2-ax+b)(x^2+ax+c) with a>0, since only this form gives zero x^3 coefficient
x^2 coefficient is -10, so b+c-a^2 = -10 and a(b-c) = -4
Look for b and c +-2,+-4 such that bc=8, with b, c having the same sign (-2,-4), (-4,-2), (+2,+4), (+4,+2)
Note that 16 - (4+2) = 10 = 4 +(4+2), and a=4 or 2
c>b since the x^2 coefficient (-ac+ab) = a*(b-c) =-10 which gives us (b,c) = (2,4) or (b,c) = (-4,-2)
(b+c-a^2) = -10, a(b-c) = -4 so a!=4, a=2, b=-4, c=-2
f(x) = (x^2 - 2x - 4)*(x^2 + 2x - 2)
Two quadratics each with real solutions since "b^2-4ac > 0" in both cases,
but I won't bother calculating the four roots, that's trivial
So with simple equations we deduce a,b,c without all the long winded unexplained mumbo-jumbo of forming two square expressions. There is nothing intuitive or formal in your method. The approach above is rigorous.
Thanks for sharing your method! I appreciate your feedback and will take it into account for future videos.
1st approach - Factorization
since there is no x^3 term,
x^4 - 10x^2 - 4x + 8 = (x^2 + ax + b)(x^2 - ax + 8/b) = x^4 + (8/b - a^2 + b)x^2 +(8a/b - ab)x + 8
=> 8/b - a^2 + b = -10 & 8a/b - ab = -4
we test for b = factors of 8, we find b = -2, a = 2 is the solution.
=> x^4 -10x^2 - 4x + 8 = (x^2 + 2x - 2)(x^2 - 2x - 4)
2nd approach - Difference of Perfect Squares
since there is no x^3 term,
x^4 - 10x^2 - 4x + 8 = (x^2 + a/2)^2 - b(x + c/2)^2 = x^4 + (a - b)x^2 - bcx + (a^2 - bc^2)/4
=> a - b = -10 & -bc = -4 & (a^2 - bc^2)/4 =8
=> solving, (a,b,c) = (-6, 4, 1)
=> x^4 - 10x^2 - 4x + 8 = (x^2 - 3)^2 - 4(x + 1/2)^2 = (x^2 - 3)^2 - (2x + 1)^2
= (x^2 - 3 + 2x + 1)(x^2 - 3 - 2x - 1) = (x^2 + 2x - 2)(x^2 - 2x - 4)
It root 5 chart < or > .7. Double CV components x has 3 difference x3 x4 conservation law proves x as 1 #app. Blind -1.76
Thanks for your feedback! I appreciate the suggestion! 😎
Eu sou professor de matemática de formação aqui no Brasil, mas não temos o costume de resolver equações com esses artifícios, portanto temos dificuldades com esse tipo de questão. (I am a mathematics teacher by training here in Brazil, but we are not used to solving equations with these devices, so we have difficulties with this type of question.)
I’m glad you found it helpful! 🔥💯Thanks for your feedback! 💖💯
Analisando com mais cuidado, percebi que o coeficiente do termo de segundo grau deve ser par, e deve ser desmembrado em duas parcelas, tais que uma seja par maior que 1 e a outra seja um quadrado perfeito. (Analyzing more carefully, I realized that the coefficient of the second degree term must be even, and must be divided into two parts, such that one is even greater than 1 and the other is a perfect square.)