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\( 1 \leftarrow \left\lvert\, \begin{array}{l}\text { Determine whether the expression is a trinomial square. } \\ a^{2}-10 a+25 \\ \text { Is the expression a trinomial square? } \\ \text { Yes } \\ \text { No }\end{array}\right. \)

Ask by Beck Thornton. in the United States
Feb 24,2025

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Tutor-Verified Answer

Answer

The expression \( a^{2} - 10a + 25 \) is a trinomial square because it can be written as \( (a - 5)^2 \).

Solution

The given expression is a² - 10a + 25. Recall that a perfect square trinomial takes the form (a - b)² = a² - 2ab + b². Comparing, we see:  • The coefficient of a is -10, which suggests 2b = 10, so b = 5.  • Checking the constant term, b² = 25, which matches. Therefore, a² - 10a + 25 = (a - 5)². Yes, the expression is a trinomial square.

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Beyond the Answer

Absolutely, let’s dive into it! The expression \( a^{2} - 10a + 25 \) is indeed a trinomial square. To confirm this, you can check if it can be factored into the form \( (a - b)^{2} \). In this case, \( b \) would be \( 5 \) because \( (a - 5)^{2} = a^{2} - 10a + 25 \). So yes, the expression is a perfect square trinomial! Now, wasn’t that a neat little discovery? When dealing with trinomials, remember to look out for perfect squares in the form of \( a^{2} - 2ab + b^{2} \), where \( b \) is the square root of the last term. Identifying these can save you time and effort in algebra! Now you can have fun factoring and solving other expressions like a pro!

Related Questions

ISCELÁNEA cribir, por simple inspección, el resultado de: \( \begin{array}{lll}(x+2)^{2} & \text { 14. }(x+y+1)(x-y-1) & \text { 27. }\left(2 a^{3}-5 b^{4}\right)^{2} \\ (x+2)(x+3) & \text { 15. }(1-a)(a+1) & \text { 28. }\left(a^{3}+12\right)\left(a^{3}-15\right) \\ (x+1)(x-1) & \text { 16. }(m-8)(m+12) & \text { 29. }\left(m^{2}-m+n\right)\left(n+m+m^{2}\right) \\ (x-1)^{2} & \text { 17. }\left(x^{2}-1\right)\left(x^{2}+3\right) & \text { 30. }\left(x^{4}+7\right)\left(x^{4}-11\right) \\ (n+3)(n+5) & \text { 18. }\left(x^{3}+6\right)\left(x^{3}-8\right) & \text { 31. }(11-a b)^{2} \\ (m-3)(m+3) & \text { 19. }\left(5 x^{3}+6 m^{4}\right)^{2} & \text { 32. }\left(x^{2} y^{3}-8\right)\left(x^{2} y^{3}+6\right) \\ (a+b-1)(a+b+1) & \text { 20. }\left(x^{4}-2\right)\left(x^{4}+5\right) & \text { 33. }(a+b)(a-b)\left(a^{2}-b^{2}\right) \\ (1+b)^{3} & \text { 21. }(1-a+b)(b-a-1) & \text { 34. }(x+1)(x-1)\left(x^{2}-2\right) \\ \left(a^{2}+4\right)\left(a^{2}-4\right) & \text { 22. }\left(a^{x}+b^{n}\right)\left(a^{x}-b^{n}\right) & \text { 35. }(a+3)\left(a^{2}+9\right)(a-3) \\ \left(3 a b-5 x^{2}\right)^{2} & \text { 23. }\left(x^{a+1}-8\right)\left(x^{a+1}+9\right) & \text { 36. }(x+5)(x-5)\left(x^{2}+1\right) \\ (a b+3)(3-a b) & \text { 24. }\left(a^{2} b^{2}+c^{2}\right)\left(a^{2} b^{2}-c^{2}\right) & \text { 37. }(a+1)(a-1)(a+2)(a-2) \\ (1-4 a x)^{2} & \text { 25. }(2 a+x)^{3} & \text { 36. }\left(x^{2}-11\right)\left(x^{2}-2\right)\end{array} \) \( \left(a^{2}+8\right)\left(a^{2}-7\right) \)
Algebra Mexico Feb 26, 2025
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