QUESTION IMAGE
Question
- $-x^2 + 25$
- $8x^2 - 98$
- $x^2 - 1$
- $-64x^2 + 121$
- $28x^2 - 7$
- $27x^2 - 12$
Step1: Factor \(-x^2 + 25\)
Rewrite as \(25 - x^2\), which is a difference of squares: \(a^2 - b^2=(a + b)(a - b)\), where \(a = 5\), \(b = x\). So \(25 - x^2=(5 + x)(5 - x)\) or \(-(x + 5)(x - 5)\).
Step2: Factor \(8x^2 - 98\)
First, factor out the GCF, which is 2: \(2(4x^2 - 49)\). Then \(4x^2 - 49\) is a difference of squares (\(a = 2x\), \(b = 7\)): \(2(2x + 7)(2x - 7)\).
Step3: Factor \(x^2 - 1\)
This is a difference of squares (\(a = x\), \(b = 1\)): \((x + 1)(x - 1)\).
Step4: Factor \(-64x^2 + 121\)
Rewrite as \(121 - 64x^2\), a difference of squares (\(a = 11\), \(b = 8x\)): \((11 + 8x)(11 - 8x)\) or \(-(8x + 11)(8x - 11)\).
Step5: Factor \(28x^2 - 7\)
Factor out the GCF, 7: \(7(4x^2 - 1)\). Then \(4x^2 - 1\) is a difference of squares (\(a = 2x\), \(b = 1\)): \(7(2x + 1)(2x - 1)\).
Step6: Factor \(27x^2 - 12\)
Factor out the GCF, 3: \(3(9x^2 - 4)\). Then \(9x^2 - 4\) is a difference of squares (\(a = 3x\), \(b = 2\)): \(3(3x + 2)(3x - 2)\).
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s:
- \(\boldsymbol{(5 + x)(5 - x)}\) (or \(\boldsymbol{-(x + 5)(x - 5)}\))
- \(\boldsymbol{2(2x + 7)(2x - 7)}\)
- \(\boldsymbol{(x + 1)(x - 1)}\)
- \(\boldsymbol{(11 + 8x)(11 - 8x)}\) (or \(\boldsymbol{-(8x + 11)(8x - 11)}\))
- \(\boldsymbol{7(2x + 1)(2x - 1)}\)
- \(\boldsymbol{3(3x + 2)(3x - 2)}\)