Factor the polynomial below 20x^2-45

Mathematics

2 answers:

stiks02 [169]3 years ago

8 0

$20x^2-45=5(4x^2-9)=5[(2x)^2-3^2]=5(2x-3)(2x+3)\\\\--------------------------\\\\use:a^2-b^2=(a-b)(a+b)$

rodikova [14]3 years ago

3 0

$(a+b)(a-b)=a^2-b^2\\---------------------\\\\20x^2-45=5\cdot4x^2-5\cdot9=5\cdot(4x^2-9)=\\\\.\ \ \ \ \ \ \ \ \ \ \ \ =5\cdot[(2x)^2-3^2]=5\cdot(2x+3)(2x-3)$

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scoundrel [369]

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Answer these 2 questions please

Whitepunk [10]

Hi there!

Question 1:

For this, we are taking away 1/4 yards of spring from 7/8 yards. This is essentially just subtracting 1/4 from 7/8. This then gives us the equation:

$\frac{7}{8}-\frac{1}{4}$

Now, to subtract these, we want to make it so the denominators are the same, or the numbers on the lower half are the same. (One way to explain why this is true is: $\frac{a}{b}-\frac{c}{b}=a\frac{1}{b}-c\frac{1}{b}=(a-c)\frac{1}{b}=\frac{a-c}{b}$ ). Now, to make it so the denominators are the same, we can multiply the second fraction, 1/4, by 2/2 as 2/2 is essentially 1, and multiplying by 1 will get the same result, just with a denominator switched in this case. Doing this, we get:

$\frac{7}{8}-\frac{1}{4}\cdot\frac{2}{2}$

$\frac{7}{8}-\frac{2}{8}$

Now, subtracting the numerators, we get:

$\frac{5}{8}$ yards of string will remain.

Question 2:

For this, we are combining these two thicknesses, and thus we add 1/4 to 7/12. This gives us the equation:

$\frac{1}{4}+\frac{7}{12}$

Now, we again want to make it so the denominators are the same. (Here's a similar proof but for addition this time: $\frac{a}{b}+\frac{c}{b}=a\frac{1}{b}+c\frac{1}{b}=(a+c)\frac{1}{b}=\frac{a+c}{b}$ ). Now, to make the denominators the same, we can multiply the first fraction, 1/4, by 3/3 (also equivalent to 1). Doing this, we get: