Question

Please help. I’ll mark you as brainliest if correct . I don’t understand this math problem. Thank you .

Answer 1

Answer:

That can be factored as

(x -1  (1/3) ) * ( x +3) * (x -4/5)

and the zeroes are located at:

x = 1.33333333...   x = -3    and x = .8

Step-by-step explanation:

Answer 2

Answer:

$\boxed{\sf \ \ \ f(x)=(x+3)(5x-4)(3x-4) \ \ \ }$

Step-by-step explanation:

We need to factorise the following function

$f(x)=15 x^3+13 x^2-80 x+48$

-3 is a trivial solution, we can notice that f(-3)=0

so we can factorise by (x+3)

let s note a, b and c real and let s write

$f(x)=15 x^3+13 x^2-80 x+48=(x+3)(ax^2+bx+c)$

$(x+3)(ax^2+bx+c) = ax^3+bx^2+cx+3ax^2+3bx+3c=ax^3+(b+3a)x^2+(3b+c)x+3c$

let s identify...

the terms in $x^3$

   15 = a

the terms in $x^2$

   13 = b + 3a

the terms in x

   -80 = 3b+c

the constant terms

   48 = 3c

so it comes, c=48/3=16, a = 15, b = 13-3*15=13-45=-32

so $f(x)=(x+3)(15x^2-32x+16)$

$\Delta=32^2-4*15*16=64$

so the roots of $(15x^2-32x+16)$ are

$\dfrac{32-8}{15*2}=\dfrac{24}{30}=\dfrac{12}{15}=\dfrac{4}{5}$

and

$\dfrac{32+8}{15*2}=\dfrac{40}{30}=\dfrac{20}{15}=\dfrac{4}{3}$

so $f(x)=(x+3)(5x-4)(3x-4)$

the zeros are -3, 4/5, 4/3