Give the algebraic expression by using variables and constant and arithmetic operations: i. Sum of the squares of x and y. ii.Pr
1 answer:
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Answer:
According to steps 2 and 4. The second-order polynomial must be added by and
to create a perfect square trinomial.
Step-by-step explanation:
Let consider a second-order polynomial of the form ,
. The procedure is presented below:
1) (Given)
2) (Compatibility with addition/Existence of additive inverse/Modulative property)
3) (Compatibility with multiplication)
4) (Compatibility with addition/Existence of additive inverse/Modulative property)
5) (Perfect square trinomial)
According to steps 2 and 4. The second-order polynomial must be added by and
to create a perfect square trinomial.
Answer:
3 + 11a³ - 7a² + 18a - 18
Step-by-step explanation:
<u>When multiplying with two brackets, you need to multiply the three terms, (a²), (4a) and (-6) from the first bracket to all the terms in the second brackets, (3a²), (-a) and (3) individually. I have put each multiplied term in a bracket so it is easier.</u>
(a² + 4a - 6) × (3a² - a + 3) =
(a² × <em>3a²</em>) + {a² × <em>(-a)</em>} + (a² × <em>3</em>) + (4a × <em>3a²</em>) + {4a × <em>(-a)</em>} + (4a × <em>3</em>) + {(-6) × <em>a²</em>) + {(-6) × <em>(-a)</em>} + {(-6) × <em>3</em>}
<u>Now we can evaluate the terms in the brackets. </u>
(a² × 3a²) + {a² × (-a)} + (a² × 3) + (4a × 3a²) + {4a × (-a)} + (4a × 3) + {(-6) × a²) + {(-6) × (-a)} + {(-6) × 3} =
3 + (-a³) + 3a² + 12a³ + (-4a²) + 12a + (-6a²) + 6a + (-18)
<u>We can open the brackets now. One plus and one minus makes a minus. </u>
3 + (-a³) + 3a² + 12a³ + (-4a²) + 12a + (-6a²) + 6a + (-18) =
3 -a³ + 3a² + 12a³ -4a² + 12a -6a² + 6a -18
<u>Evaluate like terms.</u>
3 -a³ + 3a² + 12a³ -4a² + 12a -6a² + 6a -18 = 3
+ 11a³ - 7a² + 18a - 18