3 years ago

Multiplying Binomials pls help

1 answer:

allsm [11]3 years ago

3 0

Answer:

Answer C

Step-by-step explanation:

You might be interested in

VLD [36.1K]

Answer:

(x^4) - 64x

Step-by-step explanation:

zeros at 4 and 0 are the roots so can be factor as x*(x-4) and the multiplicity tels to what power you have to raise those factors

$x^{1} *(x-4)^{3}$ becuase multiplicity of 1 for root at 0, and multiplicity of 3 at root 4

now to write this in standard form

$x*(x-4)^{3} = \\x*(x-4)(x^{2} +4x+16)= \\(x^{2} -4x)(x^{2} +4x+16) = \\x^{4} +4x^{3} +16x^{2} -4x^{3} -16x^{2} -64x = \\x^{4} -64x$

8 0

3 years ago

Crazy boy [7]

$dx+hy=j \\ \\ dx = j-hy \ \ / : \ d \\ \\d\neq 0\\ \\x=\frac{j-hy }{d}$

6 0

3 years ago

olasank [31]

Answer:

Step-by-step explanation:jdjdjsjsjsjssjjsns

8 0

3 years ago

alina1380 [7]

Step-by-step explanation:

could you please mark me as the brainliest answer

7 0

3 years ago

MrMuchimi

Answer:

Step-by-step explanation:

3 0

3 years ago