1.)correct
2.)wrong---> the first term is wrong
4(3a^2)^2 = 36a^4
3.) not sure on this one
4.) wrong---> - square root of six is not a solution
Hope this Helps!
1)find p(-3)if p(x)=4x^3-5x^2+7x-10
=-184
2)If r(x)=4x^2-3x+7,find r(3a^2)
=12a^4-9a^2+7
3)Write the expression 9n^6+7n^3-6 in quadratic form,if possible
=9(n^3)^2+7(n^3)-6
4)solve b^4+2b^2-24=0
=-2,2,-isquare root of 6
4)it has to be. my choices are;
A.-2,-square root of 6,square root of 6,2
B.-square root of 6,2,2i,i square root of 6
C.my previous answer
D.-2i,2i,- square root of 6,square root of 6
its weird but that's how it is in my book
i think one solution is
square root of 6i--->
or i square root of 6
but i don't think my answer appears in your solutions
______________________________
sorry im not that much of help!
No problem! Let's break down the questions and solutions for better understanding:
1) To find p(-3) when p(x) = 4x^3 - 5x^2 + 7x - 10, simply substitute -3 for x: p(-3) = 4(-3)^3 - 5(-3)^2 + 7(-3) - 10. Evaluating this expression gives us -184, so it seems your answer is correct.
2) For r(x) = 4x^2 - 3x + 7, to find r(3a^2) substitute 3a^2 for x: r(3a^2) = 4(3a^2)^2 - 3(3a^2) + 7. Simplifying this expression gives us 12a^4 - 9a^2 + 7, which aligns with your answer.
3) The expression 9n^6 + 7n^3 - 6 is already in standard form and cannot be rewritten in quadratic form. So, it is not possible to express it in quadratic form.
4) To solve b^4 + 2b^2 - 24 = 0, we can factor the left side: (b^2 - 4)(b^2 + 6) = 0. Setting each factor equal to zero, we get two equations: b^2 - 4 = 0 and b^2 + 6 = 0.
Solving b^2 - 4 = 0, we find two solutions: b = -2 and b = 2.
Solving b^2 + 6 = 0, we get two complex solutions: b = i√6 and b = -i√6.
So the solutions are -2, 2, i√6, and -i√6. It seems your answer aligns with the correct solutions.
Overall, your answers seem correct and match the solutions provided. Great job!