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3 years ago

8

The partial factorization of x2 – x – 12 is modeled with algebra tiles.

1 answer:

garik1379 [7]3 years ago

7 0

Answer:

x² – x – 12 = (x – 4)(x + 3)

Step-by-step explanation:

Identify two numbers that add to -1 and multiply to -12, let's call them p and q.

So ax² + bx + c = (x + p)(x + q)

pq = c

p + q = b.

It is easier to find these numbers by finding factors of -12.

This can be done by splitting the number up until all the numbers are prime.

-12 → 6 × -2 or -6 × 2 → -(3 × 2 × 2)

There can only be two numbers so the only options we have are 6 and -2, -6 and 2, 3, and -4, or -3 and 4.

We can eliminate them by adding them up.

6 + -2 = 4 ≠ -1 so that can't be it.

-6 + 2 = -4 ≠ -1 so that can't be it either.

-3 + 4 = 1 ≠ -1

therefore p and q are 3 and -4 because 3 + -4 = -1.

so x² – x – 12 = (x – 4)(x + 3)

p = -4, and q = 3.

(x – 4)(x + 3) = x(x + 3) – 4(x + 3) = x² + 3x – 4x + 12 = x² – x – 12

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3 years ago

Match the numerical expressions to their simplified forms

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Answer:

$1.\ \ p^2q = (\frac{p^5}{p^{-3}q^{-4}})^{\frac{1}{4}}$

$2.\ \ pq^{\frac{3}{2}}} = (\frac{p^2q^7}{q^{4}})^{\frac{1}{2}}$

$3.\ \ pq^2 = \frac{(pq^3)^{\frac{1}{2}}}{(pq)^{\frac{-1}{2}}}$

$4.\ \ p^2q^{\frac{1}{2}} = (p^6q^{\frac{3}{2}})^{\frac{1}{3}}$

Step-by-step explanation:

Required

Match each expression to their simplified form

1.

$(\frac{p^5}{p^{-3}q^{-4}})^{\frac{1}{4}}$

Simplify the expression in bracket by using the following law of indices;

$\frac{a^m}{a^n} = a^{m-n}$

The expression becomes

$(\frac{p^{5-(-3)}}{q^{-4}})^{\frac{1}{4}}$

$(\frac{p^{5+3}}{q^{-4}})^{\frac{1}{4}}$

$(\frac{p^8}{q^{-4}})^{\frac{1}{4}}$

Split the fraction in the bracket

$(p^8*\frac{1}{q^{-4}})^{\frac{1}{4}}$

Simplify the fraction by using the following law of indices;

$\frac{1}{a^{-m}} = a^m$

The expression becomes

$(p^8*q^4)^{\frac{1}{4}}$

Further simplify the expression in bracket by using the following law of indices;

$(ab)^m = a^m * b^m$

The expression becomes

$(p^{8*\frac{1}{4}}\ *\ q^4*^{\frac{1}{4}})$

$(p^{\frac{8}{4}}\ *\ q^{\frac{4}{4}})$

$p^2q$

Hence,

$(\frac{p^5}{p^{-3}q^{-4}})^{\frac{1}{4}} = p^2q$

2.

$(\frac{p^2q^7}{q^{4}})^{\frac{1}{2}}$

Simplify the expression in bracket by using the following law of indices;

$\frac{a^m}{a^n} = a^{m-n}$

The expression becomes

$({p^2q^{7-4}}})^{\frac{1}{2}}$

$({p^2q^3}})^{\frac{1}{2}}$

Further simplify the expression in bracket by using the following law of indices;

$(ab)^m = a^m * b^m$

The expression becomes

${p^{2*\frac{1}{2}}q^{3*\frac{1}{2}}}}$

$pq^{\frac{3}{2}}}$

Hence,

$pq^{\frac{3}{2}}} = (\frac{p^2q^7}{q^{4}})^{\frac{1}{2}}$

3.

$\frac{(pq^3)^{\frac{1}{2}}}{(pq)^{\frac{-1}{2}}}$

Simplify the numerator as thus:

$\frac{p^{\frac{1}{2}} * q^3*^{\frac{1}{2}}}{(pq)^{\frac{-1}{2}}}$

$\frac{p^{\frac{1}{2}} * q^{\frac{3}{2}}}{(pq)^{\frac{-1}{2}}}$

Simplify the denominator as thus:

$\frac{p^{\frac{1}{2}} * q^{\frac{3}{2}}}{p^{\frac{-1}{2}}q^{\frac{-1}{2}}}$

Simplify the expression in bracket by using the following law of indices;

$\frac{a^m}{a^n} = a^{m-n}$

The expression becomes

$p^{\frac{1}{2} - (\frac{-1}{2} )} * q^{\frac{3}{2} - (\frac{-1}{2}) }$

$p^{\frac{1}{2} +\frac{1}{2} } * q^{\frac{3}{2} + \frac{1}{2} }$

$p^{\frac{1+1}{2}} * q^{\frac{3+1}{2}}$

$p^{\frac{2}{2}} * q^{\frac{4}{2}}$

$pq^2$

Hence,

$pq^2 = \frac{(pq^3)^{\frac{1}{2}}}{(pq)^{\frac{-1}{2}}}$

4.

$(p^6q^{\frac{3}{2}})^{\frac{1}{3}}$

Simplify the expression in bracket by using the following law of indices;

$(ab)^m = a^m * b^m$

The expression becomes

$p^6*^{\frac{1}{3}}\ *\ q^{\frac{3}{2}}*^{\frac{1}{3}}$

$p^{\frac{6}{3}}\ *\ q^{\frac{3*1}{2*3}}$

$p^2 *\ q^{\frac{3}{6}}$

$p^2 *\ q^{\frac{1}{2}$

$p^2q^{\frac{1}{2}$

Hence

$p^2q^{\frac{1}{2}} = (p^6q^{\frac{3}{2}})^{\frac{1}{3}}$

6 0

3 years ago