QUESTION IMAGE
Question
structures of expressions - 2.10
9 a. standard form
b. vertex form
c. factored form
skip this for now
d. table
| x | y |
|---|---|
| 1 | 2 |
| 2 | - 6 |
| 3 | - 14 |
| 4 | - 16 |
| 5 | - 14 |
| 6 | - 6 |
| 7 | 2 |
| 8 | 12 |
show the first differences and the second differences.
go
topic: factoring quadratics
verify each factorization by multiplying.
- $x^{2}+12x - 64=(x + 16)(x - 4)$
- $x^{2}-64=(x + 8)(x - 8)$
- $x^{2}+20x + 64=(x + 16)(x + 4)$
- $x^{2}-16x + 64=(x - 8)(x - 8)$
factor the following quadratic expressions, if possible. (some will not factor.)
- $x^{2}-5x + 6$
- $x^{2}-7x + 6$
- $x^{2}-5x - 36$
- $m^{2}+16m + 63$
- $s^{2}-3s - 1$
- $x^{2}+7x + 2$
- $x^{2}+14x + 49$
- $x^{2}-9$
- $c^{2}+11c + 3$
- which quadratic expression above could represent the area of a square? explain.
- would any of the expressions above not be the side - lengths for a rectangle? explain.
Step1: Verify factorizations
For \(x^{2}+12x - 64=(x + 16)(x - 4)\), use FOIL method.
\((x + 16)(x - 4)=x\times x+x\times(-4)+16\times x+16\times(-4)=x^{2}-4x + 16x-64=x^{2}+12x - 64\).
For \(x^{2}-64=(x + 8)(x - 8)\), \((x + 8)(x - 8)=x^{2}-8x+8x - 64=x^{2}-64\).
For \(x^{2}+20x + 64=(x + 16)(x + 4)\), \((x + 16)(x + 4)=x^{2}+4x+16x + 64=x^{2}+20x + 64\).
For \(x^{2}-16x + 64=(x - 8)(x - 8)\), \((x - 8)(x - 8)=x^{2}-8x-8x + 64=x^{2}-16x + 64\).
Step2: Factor quadratic expressions
For \(x^{2}-5x + 6\), find two numbers that multiply to \(6\) and add to \(-5\). The numbers are \(-2\) and \(-3\), so \(x^{2}-5x + 6=(x - 2)(x - 3)\).
For \(x^{2}-7x + 6\), two numbers that multiply to \(6\) and add to \(-7\) are \(-6\) and \(-1\), so \(x^{2}-7x + 6=(x - 6)(x - 1)\).
For \(x^{2}-5x-36\), two numbers that multiply to \(-36\) and add to \(-5\) are \(-9\) and \(4\), so \(x^{2}-5x-36=(x - 9)(x + 4)\).
For \(m^{2}+16m + 63\), two numbers that multiply to \(63\) and add to \(16\) are \(7\) and \(9\), so \(m^{2}+16m + 63=(m + 7)(m + 9)\).
\(s^{2}-3s - 1\) cannot be factored over the integers.
For \(x^{2}+7x + 2\), cannot be factored over the integers.
For \(x^{2}+14x + 49=(x + 7)^{2}\) (since \((x + 7)(x + 7)=x^{2}+7x+7x + 49=x^{2}+14x + 49\)).
For \(x^{2}-9=(x + 3)(x - 3)\) (difference - of - squares formula \(a^{2}-b^{2}=(a + b)(a - b)\) with \(a=x\) and \(b = 3\)).
\(c^{2}+11c + 3\) cannot be factored over the integers.
Step3: Answer questions about geometric interpretations
For question 23, \(x^{2}+14x + 49=(x + 7)^{2}\) can represent the area of a square because the area of a square with side - length \(s\) is \(s^{2}\), and here the expression is a perfect - square trinomial.
For question 24, All quadratic expressions can potentially represent the product of the side - lengths of a rectangle (since a rectangle's area is \(A = lw\) and a quadratic expression can be written as a product of two linear expressions in some cases). If an expression cannot be factored over the real numbers, we can still consider it in the context of complex numbers or as a non - factorable quadratic representing an area in a more abstract sense. But if we are only considering real - valued side - lengths and factoring over the integers, non - factorable expressions like \(s^{2}-3s - 1\), \(x^{2}+7x + 2\), \(c^{2}+11c + 3\) would not be easily expressed as the product of two integer - valued side - lengths of a rectangle.
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- Verified.
- \((x - 2)(x - 3)\)
- \((x - 6)(x - 1)\)
- \((x - 9)(x + 4)\)
- \((m + 7)(m + 9)\)
- Cannot be factored over integers.
- Cannot be factored over integers.
- \((x + 7)^{2}\)
- \((x + 3)(x - 3)\)
- Cannot be factored over integers.
- \(x^{2}+14x + 49\) because it is a perfect - square trinomial \((x + 7)^{2}\).
- Expressions like \(s^{2}-3s - 1\), \(x^{2}+7x + 2\), \(c^{2}+11c + 3\) cannot be easily expressed as the product of two integer - valued side - lengths of a rectangle if factoring over the integers.