# If “

My initial thought is that if \$a\$, \$b\$ và \$c\$ are \$1\$ or \$-1\$, then the polynomial evaluates to lớn \$0.\$ So, maybe two of the factors will be \$(a + b + c - 3)\$ and \$(a + b + c + 3)\$. An alternative option that combines these two might be \$a^2 + b^2 + c^2 - 3\$.

Is the thought process correct here, & would trial và error be a good way lớn decide between the linear and the quadratic options I described above?As you can tell, I am largely doing guess work here. Is there a more systematic way of deciding what terms to địa chỉ and subtract in orders khổng lồ factor the polynomial?

Note: The factoring need not be done all the way lớn linear factors. All that is needed is a hàng hóa of polynomials.

Bạn đang xem: If “

I suggests that you use \$(a+b)^3=a^3+b^3+3ab(a+b)Rightarrow a^3+b^3=(a+b)^3-3ab(a+b)\$ instead, you will need to lớn use it twice lượt thích this:

\$a^3+b^3+c^3-3abc\$

\$=(a+b)^3+c^3-3ab(a+b)-3abc\$

\$=(a+b+c)^3-(3c(a+b)^2+3(a+b)c^2)-3ab(a+b+c)\$

\$=(a+b+c)^3-3c(a+b)(a+b+c)-3ab(a+b+c)\$

\$=(a+b+c)^3-(a+b+c)(3ab+3bc+3ca)\$

\$=(a+b+c)(a^2+b^2+c^2+2ab+2bc+2ca)-(a+b+c)(3ab+3bc+3ca)\$

\$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\$

Hint: divide \$\$a^3+b^3+c^3-3abc\$\$ by \$a+b+c\$ the result is given by \$\$left( c+a+b ight) left( a^2-ab-ca+b^2-bc+c^2 ight) \$\$

Thanks for contributing an answer to goodsmart.com.vnematics Stack Exchange!

But avoid

Asking for help, clarification, or responding khổng lồ other answers.Making statements based on opinion; back them up with references or personal experience.

Xem thêm: Thủy Thủ Popeye Ăn Rau Chân Vịt, Rau Chân Vịt Là Gì

Use goodsmart.com.vnJax khổng lồ format equations. goodsmart.com.vnJax reference.

Xem thêm: Hướng Dẫn Cách Cắm Hoa Giả Để Bàn Đẹp Trang Trí Nhà Thêm Lung Linh

## Not the answer you're looking for? Browse other questions tagged algebra-precalculus or ask your own question.

Spivak, Ch 2 Prologue, Problem 12d: is there a systematic way lớn find two irrational numbers whose sum and multiplication result in rational number?
For what constants \$a,b,c,d,e,f,gingoodsmart.com.vnbfZ\$ is \$(ax^2+bx+c)(fx+g)-(dx+e)=0\$ solvable via the *Rational Root Thm.*

Site thiết kế / hình ảnh © 2022 Stack Exchange Inc; user contributions licensed under cc by-sa. Rev2022.6.15.42378